Production Note
Cornell University Library pro-
duced this volume to replace the
irreparably deteriorated original.
It was scanned using Xerox soft-
ware and equipment at 600 dots
per inch resolution and com-
pressed prior to storage using
CCITT Group 4 compression. The
digital data were used to create
Cornell's replacement volume on
paper that meets the ANSI Stand-
ard Z39.48-1984. The production
of this volume was supported in
part by the Commission on Pres-
ervation and Access and the Xerox
Corporation. Digital file copy-
right by Cornell University
Library 1991.®0rntU Iftnimsitg %	Sthtatg
THE EVAN WILHELM EVANS	
MATHEMATICAL SEMINARY	LIBRARY
THE GIFT OF	
LUCIEN AUGUSTUS WAIT ■	
M.338			tlElj.io
734&-1KEY TO THE EXEKCISES
IN THE
FIKST SIX BOOKS
OP
CASEY’S ELEMENTS OF EUCLID.
SECOND EDITION.NOW READY.
Fourth Edition, Revised and Enlarged, Price 3s. 6d., Cloth.
A SEQUEL TO THE FIRST SIX BOOKS OP THE
ELEMENTS OF EUCLID,
Containing an Easy Introduction to Modern Geometry :
numerous Examples.
Fourth Edition, Price 4s. 6d.; or in two parts, each 2s. 6d.
THE ELEMENTS OF EUCLID, BOOKS I.-VL, AND
PROPOSITIONS I.-XXL, OF BOOK XI.;
Together with an Appendix on the Cylinder, Sphere,
Cone, &*c. : with
Copious Annotations & numerous Exercises.
Second Edition, Price 6s.
A KEY TO THE EXERCISES IN THE FIRST SIX
BOOKS OF CASEY’S “ELEMENTS OF EUCLID.”
Price 3s.
A TREATISE ON ELEMENTARY TRIGONOMETRY,
With Numerous Examples
AMD
Questions for Examination.
DUBLIN: HODGES, EIGGIS, & CO.
LONDON: LONGMANS, GREEN & CO.A
DEPARTMENT OF MATHEMATI63
COR NEIL UNIVERSITY
KEY TO THE EXERCISES
IN THE
FIEST SIX BOOKS
OF
CASEY'S ELEMENTS OF EUCLID.
BY
JOSEPH B. CASEY,
•ί	’
TUTOR, UNIVERSITY COLLEGE, DUBLIN.
SECOND EDITION, REVISED.
DUBLIN: HODGES, FIGGIS, & CO., GRAFTON-ST.
LONDON: LONGMANS, GREEN, & CO.
1887.DUBLIN :
PRINTED AT THE UNIVERSITY PRESS»
BY PONSONBY AND WELDRICK.CONTENTS.
PAGES
BOOK 1.........................................1-52
Propositions.
i. 1 ; ii. 2 ; iv. 3; y. 3 ; vii. 5 ; ix. 6 ; x. 6 ; xi. 7;
xii. 8; xvii. 9; xvm. 9; xix. 10; xx. 11; xxi. 12;
xxiii. 13; xxiv. 14; xxv. 14; xxvi. 15 ; xxix. 17;
xxxi. 18; xxxii. 21 ; xxxm. 24; xxxiv. 25;
xxxvi. 26; xxxvn. 27; xxxviii. 28; xl. 29;
xlv. 31 ; xlvi. 31 ; xlvii. 33; Miscellaneous
Exercises on Book I. 36.
BOOK II.
53-70
Propositions.
iv. 53; v. 54; vi. 55; viii. 56; ix. 57; x. 58; xi. 59;
xii. 61; xiii. 62 ; xiv. 63; Miscellaneous Exercises
on Book II. 63.
BOOK III.
71-138
Propositions.
hi. 71; xiii. 73; xiv. 74; xv. 75; xvi. 76; xvii. 79
xxi. 82; xxii. 84; xxvm. 88 ; xxx. 90 ; xxxii. 91
xxxm. 93; xxxv. 98; xxxvi. 102; xxxvn. 102
Miscellaneous Exercises on Book III. 105.Vi	CONTENTS.
AGES*
BOOK IY.................................139-183
Propositions.
iv. 139; v. 143; x. 144; xx. 145; xv. 146; Exer-
cises on Book IY. 148.
BOOK Y.........................................184-186
Miscellaneous Exercises, 183.
BOOK VI...................................... 187-282
Propositions.
n. 186 ; m. 186 ; it. 188 ; x. 190; xi. 193; xm. 193;
xvii. 196; xix. 200; xx. 200; xxi. 204; xxm. 205;
xxx. 205; xxxi. 207 ; Exercises on Book YI. 207.EXERCISES ON EUCLID,
—♦—
BOOK I.
PROPOSITION I.
1. Dem.—The four lines AC, AF, BC, BF are each = AB, and
.·. = to each other. Hence ACBF is a lozenge.
2.	Bern.—Because AC = BC, and CF common, and the base
AF = BF; .·. (viii.) the L ACF = BCF; .·. ACF is J an L of
an equilateral Δ. Again, the L CAB = ACD + ADC (xxxii.) ;
but ACD = ADC; .·. CAB = 2ACD; .·. ACD is J an L of an
equilateral Δ, and ACF is £ an L of an equilateral Δ; .*. DCF
is an L of an equilateral Δ. Similarly DFC is an L of an equi-
lateral Δ. Hence the Δ CDF is equilateral.
3.	Bern.—Join AF. Because AG = AF, the L AGF = AFG;
and because AF = AC, the L ACF = AFC; .*. the L GFC = FGC
+ FCG, and is .·. (xxxii. Cor. 7) a right L, In like manner
HFC is a right L. Hence (xiv.) G, F, H are collinear.
4.	Bern.—GC2 = GF2 + FC2 (xlvii.), and GC2 = 4AG2;
.·. GF2 + FC2 = 4 AG2; but GF = AG. Therefore FC* = 3AG2
= 3AB2.
5.	Sol.—Join CF. Divide AD into four equal parts in E, G, H.
From DC cut off DJ = ED. J is the centre of the required Θ.
Bern.—Join AJ, BJ, and produce them to meet the Θ* in K, L.
B2
EXERCISES ON EUCLID.
[book I.
Because the L ADJ is right, AJ2 = JD2 + DA2 = 3* + 42 = 52;
AJ is = 6 of the parts into which AD is divided; tpit AK
= AB; .·. JK = 3 of the parts ; JK as JD. Again, AD = DB,
and DJ common, and the L ADJ equal BDJ;	(iv.) AJ = BJ;
but AK = BL; .·. JK « JL. Hence the lines JD, JK, JL are
equal; and the Θ, with J as centre and JD as radius, will pass
through the points K, L.
PROPOSITION II.
1. Sol.—On AB describe the equilateral Δ ABD. With B as
centre and BC as radius, describe the Θ CEG, and produce DB
to meet it in E. With D as centre and DE as radius, describe
the Θ EFH, and produce DA to meet it in F. AF is the re·
quired line.
Bern.—Because D is the centre of the Θ EFH, DE = DF;
but DB = DA; .·. BE = AF, and BE = BC; AF = BC.
2. Sol.—Let A be the given point, and BC the given line.
It is required from the point A to inflect to BC a line equal to a
given line DE. From A draw AF = DE [n.]. With A as centre,BOOK I.]
EXERCISES ON EUCLID
3
and AF as radius, describe a Θ cutting BC in G H. Join AG,
AH. AG, AH are the required lines. ,
Dem.—Because AF = AG, and AF = DE; .·. AG = BE. In
like manner AH = DE. Hence there are two solutions.
PROPOSITION IY.
1.	Let AD bisect the vertical L of the isosceles Δ ABC. It
is required to prove that it bisects the base BC perpendicularly.
Dem.—AB = AC, and AD common, and the L BAD = CAD
.·. (iv.) the L ADB = ADC, and the side BD = CD. Hence BC
is bisected, and (Def. xiv.) AD is JL to BC.
2.	Dem.—Let ABCD be the quadrilateral, and BD its diago-
nal. Because AB = CB, and BD common, and the L ABD = CBD ;
,·. (rr.)the base AD = CD.
3.	Let the lines AB, CD, bisect each other in E.
Dem.—Take any point F in ED. Join AF, BF. Because
AE = BE, and EF common, and the L AEF = BEF; .·. the
base AF = BF.
4.	Let ABC be the Δ. On the sides AB, AC, describe equi-
lateral Δ8 ABD, ACE. Join CD, BE. It is required to prove
that CD = BE.
Dem.—Because the L DAB = CAE, to each add the L BAC ;
then the L DAC = BAE; and since DA = BA, and CA = EA, the
sides DA, AC = BA, AE, and we have shown that the L DAC
= BAE; (iv.) the bases CD, BE, are equal.
PROPOSITION Y.
1. (1) Dem.—Take any point D in AB, and from AC cut off
AE = AD (in). Join BE, CD, DE. Because AB = AC, and
AE = AD ; BA and AE = CA and AD, and the L A is com-
mon; .*. BE = CD, and the L ABE = ACD. Again, because
BE = CD, and BD = CE; .·. BD and BE = CE and CD, and
the L DBE = ECD; .*. (iv.) the L BDE = CED, and the L BED
= CDE; hence the remainders, the L * BDC, BEC, are equal.
Again, BD = CE; and DC = EB; .·. BD and DC = CE and EB,
and the contained L8 BDC, CEB, have been shown to be equal;
.·. (iv.) the L8 DBC, ECB, are equal.4
EXERCISES ON EUCLID.
[book I-
(2) Dem,—ProduCe BA, CA, to J, H, in AJ; take any point»
E, G, and from AH cut off AD = AE, and AF = AG. Join DGr
DB, EC, EF. Because AF = AG, and AE = AD; .·. AF an#
AE = AG and AD, and the L FAG common; . *. the base FE = DGr
and the L AFE = AGD, and the L FEA = GDA.
Again, because BG = CF, and GD = FE; .·. BG and GD
= CF and FE, and the L DGB = EFC; .·. the base DB = EC,,
and the L GDB = FEC; hut the L GDA = FEA; .·. the re-
mainders, the L · BDC, BEC, are equal.
Now, since BD = CE, and DC = EB ; .·. BD and DC = CE
and EB, and the L BDC = CEB; the L DCB = EBC.
2. Dem.—If AH be not an axis*of symmetry, let AJ be one-
Join JG. Because AF = AG, and AJ common, and the L FAJ
GAJ (hyp.); .·. the L AFJ = AGJ; but the L AFC = AGB ;
the L AGJ = AGB, a part = to the whole, which is absurd ;
AH must be an axis of symmetry.BOOK I.]
EXERCISES ON EUCLID.
5
3.	Let ABC, DBG, be the two isosceles Δ* on the same base.
Join their vertices A, D.
Dem.—The L ABC = ACB (v.), and the L DBC = DCB (v.);
.·. the L ABD = ACD. Now, the two Δ8 ABD, ACD, have the
two sides AB, BD = the two sides AC, CD, and the contained Ls
ABD, ACD, equal; .·. (iv.) the L BAD = CAD. Hence AD is
an axis of symmetry.
4.	Let AD be the bisector of the L BAC.
Dem.—Because BA and AD = CA and AD, and the L BAD
* CAD (hyp.); .·. (iv.) the L ABD = ACD.
5.	Let ABCD be the lozenge, and AD, BC, its diagonals.
Dem.—Because AB = AC, the L ACB = ABC, and because
DB = DC, the L DCB = DBC; .*. the L ACD = ABD. Now,
the Δ8 ACD, ABD, have two sides AC, CD, and the contained
L ACD, equal to the sides AB, BD, and the contained L ABD ;
.·. (iv.) the L CAD = BAD, and the L CDA = BDA. Hence
AD is an «-xia of symmetry.
6.	Let ABC be the Δ.
Dem.—Take three points D, E, F, in the sides AB, BC, CA,
equally distant from the vertices A, B, C. Join DE, EF, FD.
It is required to prove that the Δ DEF is equilateral. Evidently
from the given conditions the Δ8 BDE, CEF, AFD, are equal;
their bases DE, EF, FD, are equal. Hence the Δ DEF is
equilateral.
PROPOSITION VII.
3. If possible let two Θ8 whose centres are A, B, intersect in
'the points C, D, on the same side of the line AB.
Dem.—Join CA, DA, CB, DB. Because A is the centre of
the Θ ECD, AC = AD; and because B is the centre of the6	EXERCISES ON EUCLID.	[book i-
Θ FCD, BC = BD; but this is contrary to Prop. vii. Hence
the Θ* cannot intersect in more than one point on the same side
of the line AB.
PROPOSITION IX.
3.	Dem.—Because AD = AE, the L AED = ADE; and be-
cause FE as FD, the L FDE =* FED. Now we have two Δ*
ADF, AEF, having two sides AD, DF, and the contained L ADF
respectively = to the two sides AE, EF, and the contained L AEF ~r
(iv.) the L DAF = EAF.
4.	Dem.—Let G be the point where AF meets DE. Because
AD = AE, and AG common, and the L DAG = EAG; .·. the
L AGD = AGE. Hence (Def. xiv.) AF is X to DE.
5.	See Ex. 3, Prop. iv.
6.	Dem.—Take any point G in AF, and from G let fall the i.
GH on AB. From AC cut off AJ = AH, and join GJ. Because
AH == AJ, and AG common, and the L HAG = JAG; .·. (rv.) the
L AJG = AHG. Hence the L AJG is right, and the base
GH = GJ.
PROPOSITION X.
1. Sol.—Let AB be the given line. Take a part AE greater
than half AB. With A as centre and AE as radius, describe the
Θ CED. Take BF = AE. With B as centre and BF as radiusr
describe the Θ CFD, cutting the Θ CED in C, D. Join CD,,
cutting AB in G. AB is bisected in G.BOOK I.]
EXERCISES ON EUCLID.
7
Dam.—Join AC, BC, AD, BD. Because AC * BC, and CD
common, and the base AD = BD;	(vm.) the L ACD = BCD.
Again, since AC = BC, and CG common, and the L ACG = BCG;
.·. (iv.) AG = BG.
2.	Dem.—Take any point H equally distant from A, B. Join
AH, BH, CQ. Because AC = BC, and CH common, and the
base AH = BH; .·. (vm.) the L ACH = BCH. Hence any
point equally distant from A, B, is in the bisector of the L ACB.
PROPOSITION XI.
1.	Hem.—Let the diagonals AD, BC, of the lozenge ABCD,
intersect in E. Because AB = AC, and AD common, and the
base BD = CD;	(vm.) the L BAE = CAE. Again, AB = AC,
AE common, and the L BAE = CAE; (iv.) BE = CE, and
the L AEB = AEC. Hence AD bisects BC perpendicularly.
2.	Dem.—Because DF = EF, the L FED = FDE (v.), and
CD = CE; .·. (iv.) the Δ DCF = ECF; .·. the L DCF = ECF,
and (Def. xiv.) each of them is a right angle.
3.	Sol.—Let AB be the given line. At the point A draw AC,
making an angle with AB. In AC take AD = AB. At D erect
DE JL to AC. Bisect the L BAC by AE, meeting DE in E.
Join BE. BE is J. to AB.
Dem.—AD = AB, AE common, and the L DAE = BAE;
(iv.) the L ADE = ABE; but ADE is a right angle (conet.);
hence ABE is a right angle.
4.	Sol.—Let AB be the given line, and C, D, the points. Join
CD; bisect CD in E. Draw EF JL to CD, meeting AB in F. F
is the required point.8
EXERCISES ON EUCLID.
[book I.
Dem.—Join CF, DF. Because (iv.) the Δ CEF = DEF;
FC = FD. Hence the point F is equally distant from C
and D.
5.	Sol.—Let AB be the given line, and C, D, the points.
From C let fall a X CG on AB, and produce it to E, so that GE
will be equal to CG. Join ED, and produce it to meet AB in F.
F is the required point.
Dem.—Join CF. Because CG = EG, and GF common, and
the L CGF = EGF; (iv.) the L CFG = EFG. Hence the
L CFD is bisected by the line AB.
6.	Sol.—Let A, B, C, be the three given points. Join AB,
BC. Bisect AB at D, and erect DF X to AB. Bisect BC at E,
and erect EF X to BC. F is the required point.
Dem.—Join AF, BF, CF. Because AD = BD, and DF com-
mon, and the L ADF = BDF; (it.) AF = BF. In like
manner BF = CF. Hence the three lines AF, BF, CF, are
equal.
PROPOSITION XII.
1. Dem.—If poeeible let FGJK be a O meeting AB in the
{K)ints F, G, J, E. Bisect FG in H. Join CH, and produce it to
M. Join CF, CG. Bisect GJ in N. Join CN, CJ, CE. Be-
cause FH = GH, and HC common, and the base FC = CG;
.·. the L FHC = GHC, and (Def. xiv.) each of them is a right
angle.
Again, since GN = JN, and CN common, and the base CG
= CJ; the L CNG = CNJ, and each is a right angle. Hence
the L CNH = CHN; .*. CH = CN; but CN is greater than CE,
because the point N is outside the Θ;	CH is greater than CE,
and CM = CE; .*. CH is greater than CM, which is absurd.
Hence the Θ cannot meet AB in more than two points.EXERCISES ON EUCLID.
9
BOOK I.]
2.	Dem.—Let ABC be the Δ, having the L BAC equal to the
sum of the L ■ ABC, ACB. Bisect AB in D, and erect DE _L to
AB, meeting BC in E. Join AE.
Because AD = BD, DE common, and the L ADE = BDE;
„·. (iv.) the L DAE = DBE; but the L BAC = ABC 4- ACB;
hence the L EAC = ECA; .·. each of the Δ8 ABE, ACE, is
isosceles; and since AE = BE = CE; .*. BC = 2AE.
PROPOSITION XVII.
Dem.—Let ABC be the Δ. Take any point D in BC. Join
AD. The L ADC is greater than ABC (xvi.), and the L ADB is
greater than ACB; but ADC and ADB equal two right angles;
«*. ABC and ACB are less than two right angles.
PROPOSITION XVIII.
1.	Dem.—Let ABC be the Δ, of which AC is greater than
AB. From AC cut off AD = AB. With A as centre, and AB as
radius, describe the circle ADE, cutting CD produced in E.
Join AE. Now the L ABC is greater than AEB; but AEB
= ABE; .*. ABC is greater than ABE, and ABE is greater than
ACB (xvi.). Hence ABC is greater than ACB.
2.	Dem.—Produce AB to D, so that AD = AC. Join CD.
Now the L ABC is greater than ADC (xvi.); but ADC = ACD;
.·. ABC is greater than ACD. Much more is ABC greater than
ACB.
3.	Dem.—Let ABCD be a quadrilateral, whose sides AB, CD,
are the greatest and least. It is required to prove that the
L ADC is greater than ABC. Join BD. Because BC is greater
than DC, the L BDC is greater than DBC (xvm.). Similarly the
L ADB is greater than ABD. Hence the L ADC is greater than
ABC.
4.	Dem.—Let ABC be a Δ, whose side BC is not less than
AB or AC. From A let fall a _L AD on BC. Because BC is
not less than AB, the L BAC is not less than BCA; .*. BCA
must be acute. In like manner CBA must be acute. Hence
AD must fall within the Δ ABC.10
EXEBCISES ON EUCLID.
[book I-
PEOPOSITION XIX.
1.	Dem.—Bisect BC in D. Join AD; produce it to E, so
that DE = AD. Join BE. Now the Δ· BDE, ADC, have the
«idee BD, DE, of one respectively equal to CD, DA, of the other,
and the contained Ls equal (xv.); .·. (iv.) BE = AC, and the
L DBE = DCA; hut the L ABD is greater than DCA (hyp.);
.*. ABD is greater than EBD; hence the line BF which bisects
the L ABE falls above BC. Produce BF to G, and make GF
= BF. Now, since ED = AD, EF is greater than AF. Cut off
FH = AF. Join GH, and produce it to meet BE in I. Now we
have in the Δ8 AFB, GFH, two sides AF, FB, in one equal HF,
FG, in the other, and the contained L 8 equal; hence AB = GH,
and the L ABF = HGF; but ABF = FBI (const.); .·. BGI
- GBI, and .·. (v.) IB = IG; but EB is greater that IB, and IG
greater than HG; .*. EB is greater than GH, and we have proved
BE = AC, and GH = AB. Hence AC is greater than AB.
2.	Dem.—Take any point D in the base BC of an isosceles
Δ ABC. Join AD. Now the L ADC is greater than ABD
(xvi.), and .*. greater than ACD. Hence (xix.) AC is greater
than AD.
If we take the point D in the base produced, we have the
L ACB, that is, ABC greater than ADC; .·. AD is greater
than AB.
3.	Dem.—This follows from the last exercise. For when we
took the point in the base, and joined it to the vertex, the joining
line was less than either side of the triangle; and when the point
was in the base produced, the joining line was greater.BOOK I.]
EXERCISES ON EUCLID.
IE
4.	(1) Dem.—Let A be the given point, and EF the given line.
From A let fall a 1 AB, and draw any other line AC to EF.
The L ACB is less than ABC (xvii) ; .·. (xix.) AC is greater
than AB.
(2) Dem.—Take another point D in EF. Join AD. Now
the L ACD is greater than ABC, and therefore obtuse; hence
ADC must be acute; AD is greater than AC.
5.	Dem.—Because AB is greater than AC, the L ACB is
greater than ABC (xviii.). Much more is the L BCF greater
than CBF. Hence (xix.) BF is greater than CF. Again (hyp.),.
AB is greater than BC; but AB = CF (iv.); .·. CF is greater
than BC; (xvin.) the L CBF is greater than CFB, that is,,
than ABE. Hence ABE or CFB is less than half ABC.
PROPOSITION XX.
1.	Dem.—Let ABC be a A. It is required to prove that the
difference between two sides AB, AC, is less than BC. From AC
cut off AD = AB, and join BD. Now AB and BC are greater
than AD and DC; but AB = AD;	BC is greater than DC, that
is, greater than the difference between AB and AC.
2.	Dem.—Let D be any point within a A ABC. Join AD,
BD, CD. Now (xx.) DA + DB > AB; DB + DC > BC; DC
+ DA > AC. Adding, we get 2 (DA + DB + DC) > (AB + BC
+ CA); .·. (DA + DB + DC) > (AB * BC + CA).
3.	Dem.—Let AD be the bisector of the L BAC. Take any
point E in AD. Join BE, CE. From AB cut off AF = AC, and.
join EF. Because AF = AC, and AE common, and the L EAF
= EAC; .·. (iv.) the base EF = EC. Again, since EF = EC,.
the difference between BE and EC is equal to the difference be-
tween BE and EF; but BE — EF is less than BF (Ex. 1) y
BE — EC is less than BF; but BF is the difference between.
BA and AC. Hence the difference between BE and EC is less
than the difference between BA and AC.
4.	Dem.—Produce BA to F, so that AF = AC. Take any point
E in the external bisector AD. Join EB, EC, EF. Now (iv.)
EF = EC. To each add EB, and we have EF and EB = EC and
EB; but EF and EB are greater than FB, that is, greater than
AB and AC. Hence EB and EC are greater than AB and AC..12
EXERCISES ON EUCLID.
[book i.
5.	Dem.—Let ABCD be the polygon. Join BD. Now (xx.)
AB 4 AD > BD; and BC 4 BD > CD ; hence AB 4 AD 4
BC > CD.
6.	Dem.—Let the Δ DEF be inscribed in ABC. Now (xx.)
AD 4 AE > DE; EC + CF>EF; FB 4 BD > FD. Adding,
we get (AB 4 BC 4 CA) > (DE 4 EF 4 FD).
7.	Dem.—Let the polygon FGHJE be inscribed in the poly
gon ABODE. Now (xx.) AF + AG > FG; BG + BH > GH;
CH 4 CJ > HJ) DJ 4 DE > JE; EE + EF > EF. Adding,
we get the perimeter oi ABODE greater than that of FGHJE.
8.	Dem.—Let ABCD be a quadrilateral, AC, BD, its diago-
nals. Now, if AC, BD, are not equal, one of them must be the
greater. Let BD be the greater; then we have the sum of the
sides AB, BC, CD, DA, greater than 2BD, and greater than
AC and BD.
9.	Dem.—Let ABC be the Δ, AD one of its medians. Pro-
duce AD to E, so that ED = AD. Join EC. Now (iv.) EC = AB,
and (xx.) AC and CE, that is, AC and AB, are greater than AE,
that is, greater than 2AD. Similarly BC and CA are greater
.than 2CG, and AB and BC are greater than 2BF; (AB 4 BC
4 CA) > (AD 4 BF 4 CG).
10.	Dem.—Let the diagonals AC, BD, of the quadrilateral
ABCD intersect in E. Take any other point F in the quadrila-
teral. Join AF, BF, CF, DF. Now (xx.) BF 4 FD > BD,
-and AF + FC > AC. Adding, we get (AF 4 BF 4 CF 4 DF)
> (AC 4 BD).
PROPOSITION XXI.
1.	Dem.—Let ABC be the Δ, and 0 any point within it.
-Join OA, OB, OC. Now, AB 4 AC > OB 4 OC (xxi.); AC 4 BC
>	OA 4 OB; and AB 4 BC > OA 4 OC. Adding, we get
2(AB 4 BC 4 CA) > 2(OA 4 OB 4 OC); .·. (OA 4 OB 4 OC)
>	(AB 4 BC 4 CA).
2.	Dem.—Produce BC both ways to meet AM, DN, in E, F.
Now (xx.) AE 4 EB > AB, and DF 4 FC > DC. To each add
BC, and we have AE 4 EF 4 FD > AB 4 BC 4 CD. Again, EM
4 MN 4 NF > EF (Ex. 5, xx.). To each add AE and DF, and we
get AM 4 MN 4 ND > AE 4 EF 4 FD; but we have shown
that AE 4 EF 4 FD > AB 4 BC 4 CD; AM 4 MN 4 ND > AB
4- BC 4 CD.JfOOK 1.]
EXERCISES ON EUCLID.
13*
PROPOSITION XXIII.
1.	Sol.—Let A, B, be the given sides, and C the L between
them. Draw any line DG, and from DG cut off DE = A. At
the point D in DG draw DH, making the L GDH = C (xxiii.).
In DH take DF = B, and join EF. DEF is the Δ required.
2.	Sol.—Let AB he the given side, and D, E, the given angles.
At the point A in AB make the L BAC = D, and at the point B in
AB make the L ABC = E. ABC is the Δ required.
3.	Sol.—Let A, B, he the given sides, and C the given angle,
Draw any line DG, and in it make DE = A, and EF = B. At
the point D in DG make the L GDH = C. With E as centre,.
and EF as radius, describe a Θ, cutting DH in J, K. Join EKr.
EJ. Then evidently either of the Δ8 DEJ, DEK, will fulfil the
given conditions.
4.	(1) Sol.—Let AB he the base, C the given L, and S the
sum of the sides. At the point A in AB make the L BAF = C,
and in AF take AE = S. Join BE. At the point B in BE make
the L EBG = BEG. ABG is the Δ required.
Dem.—Because the L EBG = BEG; .*. (vi.) EG = BG. To
each add AG, and we have AG + GB = AE; hut AE = S (const.) ;
.·. AG + GB = S.
(2) Sol.—Let AB he the base, C the given L, and D the diffe-
rence of the sides. At the point A in AB make the L BAG = C,
and let AG = D. Produce AG to E. Join BG, and at the point
B in BG make the L GBE = EGB. AEB is the Δ required.
Ddm.—Because the L GBE = EGB; .*. (vi.) EG = EB; hut
AE - GE = AG; .·. AE - BE = AG = D. Hence the difference
between AE and BE is D,14
EXERCISES ON EUCLID.
[book r.
5. (1) Let A, B, be two points, one of which, B, is in the given
line GF. It is required to find another point C in GF, s\Lch that
OB + CA may be equal to a given line D.
Sol.—In GF take a part BE = D. Join AE, and at the point
A in AE make the L CAE = CEA; then C is the required point.
Dem.—Because the L CAE « CEA, CA = CE (vi.). To each
add CB, then CA 4* CB — BE { but BE = D | CA + CB = D.
Hence C is the required point.
(2) Let A, B, be the points, GF the given line.
Sol.—In GF take a part BG = D. Join AG, and at the point
A in AG make the L GAE = AGE. E is the required point.
Dem.—Because the L GAE = AGE, GE = AE ; .·. AE — EB
= GE — EB; but GE — EB = GB ,* that is, equal to D. Hence
AE - EB = D.
1.	Dem.—At the point A, in AB, make the L BAH = EDF,
and make AH = AC or DF. Join BH. Now (iv.) BH = EF.
And because the L BAC is greater than EDF, the bisector of the
L HAC must fall to the right of AB. Let AG be the bisector.
Join HG. Now since AH = AC, and AG common, and the
L HAG = CAG; .·. (iv.) GH = GC. To each add BG, and we
have BC = HG + GB; .·. (xx.) BC is greater than BH; that is,
greater than EF.
2.	Dem. (Diagram to Ex. 1).—The L AHG = ACG; but AHG
is greater than AHB; .·. ACG is greater than AHB; that is,
greater than EFD.
1. Dem.—From BC cut off BH = EF. On BH describe the
Δ BGH = DEF; that is, having BG *= DE, and GH = DF. Join
PROPOSITION XXIY.
PROPOSITION XXY.
kF.IBOOK I.]
EXERCISES ON EUCLID.
15
AG. Because BA ■= DE, and BG = DE; .*. BA=BG; .·. (τι.)
the L BGA = BAG. Produce GH to meet AC in K. Now since
AC = DF, and GH = DF; .\ AC = GH; .·. GK is > AK;
-*. (xvm.) the L GAK is > AGK; but BAG = BGA; BAC is
> BGH, that is, > EDF.
PROPOSITION XXVI.
1.	Let ABC be the triangle.
Dem.—Let fall the 1 AD on BC. Now (xxvi.) the Δ· ADB,
ADC, are equal; DB = DC. Take any point E in AD. Join
BE, CE. Now (iv.) the ASBDE, CDE, are equal;	BE = CE.
Hence the point E is equally distant from the points A, B.
2.	Let AD bisect the vertical L BAC, and also the base BC.
Dem.—Produce AD to E, so that DE = AD. Join EC. Now
(iv.) the Δ· ADB, EDC, are equal; .*. AB = CE, and the
Z.	BAD = CED; but BAD = CAD (hyp.); .·. CAD = CED;
hence (τι.) CE = CA; but CE = BA; .·. CA = BA. Hence the
Δ BAC is isosceles.
3.	Let AB, AC, be two fixed lines, and D a point equally dis-
tant from them.
Dem.—Let fall X» DE, DF, on AB, AC. Join EF, AD.
Because DE = DF, the L DFE = DEF; but the L DFA = DEA;
-·. the L AFE = AEF, and .·. AE = AF. Now AE = AF, AD
common, and the base DE = DF; .*. the L EAD = FAD; .*. the
bisector of the L BAC is the locus of the point D. In like
manner, if we produce BA to G, the locus of a point equally dis-
tant from AC, AG, will be the bisector of the L CAG.
4.	Let AB be the given right line, and CD, EF, the other
lines.
Sol.—Let CD, EF, intersect in G, and meet AB in H, J.
Bisect the L HGJ by GK, meeting AB in K. K is the point
required.
Dem.—Let fall X· KM, KN, on CD, EF. Because the
L NGK = MGK, and GNK = GMK, and GK common;
(xxvi.) KN = KM.
5.	Let ABC, DEF, be two right-angled Δβ, having the base
BC = EF, and the acute L ABC = DEF.
Defei.—The Δ· ABC, DEF, have the L · BAC, ABC, equal
to the L · EDF, DEF, and the side BC = EF; .·. (xxvi.) they
are equal in every respect.16
[book i.
EXEKGISES ON EUCLID.
6.	Let the right-angled Δ8 ABC, DEF, have the sides AB, DE,
equal, and also their hypotenuses BC, EF equal. It is required to
prove that the Δ· are equal in every respect.
Dem.—At the point B in BC make the L GBC = DEF (xxin.),
and make BG = DE or AB. Join CG, AG.
Now the Δ8 GBC, DEF, have the sides GB, BC = DE, EF,
and the L GBC = DEF; .·. (it.) CG = DF, and the L BGG
= EDF; but EDF is a light L; .·. BGC is right, and .*. = BAC.
Now BG = DE, and DE = AB; .*. BG = AB; .·. the L BAG
= BGA; but BAC = BGC; .·. CAG = CGA; hence CG = CA;
but CG = DF; .·. AC = DF. Hence the Δ· ABC, DEF, are
equal in every respect.
7.	Let ABC be the Δ, and let the bisectors of the Δ* ABC,
ACB, meet in 0. Join OA. It is required to prove that OX
bisects the L BAC.
Bern.—From 0 let fall X· OD, OE, OF, on AB, BC, CA.
Join DF. The Δ· OBD, OBE, are equal (xxvi.); OD = OE.
Similarly OE = OF; .*. OF = OD, and .·. (v.) the L ODF = OFD;
hut the L ODA = OFA (const.); the L ADF = AFD;
.*. (vr.) AF = AD. Now AF = AD, AO common, and the base
OF = OD; hence (viii.) the L OAF = OAD. Therefore AO is
the bisector of the L BAC.
8.	Let ABC be the Δ, and let BO, CO, bisecting the two ex-
ternal L ■ meet in 0. Join OA. It is required to prove that
OA bisects the L BAC.
Dem.—From 0 let fall X* OD, OE, OF, on AB, BC, CA.
Join DF. Now, as m the last Exercise, OD = OF; .·. the L OFD
= ODF; but the L OFA = ODA; .·. AFD = ADF, and .·. AD
— AF. Now AD = AF, AO common, and the base OD = OF;
.·. the L OAD = OAF. Therefore AO bisects the L BAC.
9.	Let A, B, C, be the given points. It is required to draw a
line through C, such that the Xs on it from A, B, may be equal.
Sol.—Join AB; bisect it in 0. Join CO, and produce it to D.
From A, B, let fall the X* AE, BF, on CD.
Dem.—Because AO = BO, and the Xs AEO, AOE=BFO, BOF;
(xxvi.) AE = BF.
10.	Let AB, AC, be the given lines, and D the given point.
Sol.—Bisect the L BAC by AE. From D let fall a X DE on
AE, and produce it both ways to meet AB, AC, in B, C.
Dem.—The Δ· ABE, ACE, have the L· AEB, EAB, equal to
the L* AEC, EAC, and the side AE common; the L ABEBOOK I.]
'· EXERCISES ON EUCLID·
17
=* ACE. Hence the Δ ABC is isosceles. There are two solu-
tions. For if we produce BA to F, bisect the L CAF by AG,
and from D let fall the 1 DH on AG, and produce it to meet BA
in F, we will have another isosceles triangle.
1.	(1) Bern.—If AB, CD, are not parallel, let them meet in E.
Then we have the exterior L EGK of the A GKH equal to the
interior L GHK; but this is impossible (xvi.). Therefore AB,
CD, must be parallel.
(2) If AB, CD, are not parallel, let them meet in K, Then we
hare the L%KGH, GHK, of the A GHK, equal to two right
angles, which is impossible (χγπ.)· Hence AB, CD, must be
parallel.
2.	Let AB, CD, be the || lines, and AC,BD,the X· intercepted
between them.
Dem.—Join AD. Now, the L ACD is right (hyp.), and ABD,
CDB, together equal two right L · (xxix.); but CDB is right;
.·. ABD is right, and hence = ACD, and the L BAD = ADC
(xxix.). Therefore the Δ» ABD, ACD, have two L· of one
equal to two L · of the other, and the side AD common. Hence
(xxvi.) BD = AC.
3.	Let EF be || to AB.
Dem. Bisect the L ACD, BCD, by CE, CF. Now (xxix.)
E D F
the L ACE = DEC; but ACE = DCE; .·. DEC = DCE, and
,% DC « DE. In like manner DC = DF. Therefore DE ar DF.
PROPOSITION XXIX.
C18	. EXERCISES ON EUCLID.	[book i.
4.	Let EF be the line whose middle point is 0, and terminated
by the parallels AB, CD
Dem.—Through 0 draw a line OH, meeting ΔΒ, CD in 0, H.
The L GOE = HOF (xv.), and the L GEO = OFH (xxix.), and
OE = OF (hyp.); therefore (xxvi.) OG = OH.
5.	Let AB, CD, be the ||% and 0 the point equidistant from
them.
Dem.—Through 0 drawEF, meeting AB, CD, in E, F, and draw
GH, JK, meeting them in G, H, J, K. Because EF is bisected
in 0, .·. (4) GH, JK, are bisected in 0; then the Δβ GOJ, HOK,
have two sides GO, OJ, and the L GOJ in one equal to the sides
HO, OK, and the L HOK in the other. Hence (iy.) GJ = HK.
6.	Let AEFD be-the parallelogram formed by drawing parallel
lines from a point F in BC to the sides AB, AC, of the equi-
lateral Δ ABC.
Dem.—The L EFB = ACB (xxix.); . ·. EFB is an L of an equi-
lateral Δ, and EBF is an L of an equilateral Δ (hyp.); EBF
is an equilateral Δ; .·. EF = BF; but EF = AD; .·. EF + AD
= 2BF. In like manner, AE + DF = 2CF. Hence AE + AD
+ FE ■+* FD =- 2BC.
7.	Let ABCDEF be the hexagon, and let its diagonals AD, BE
intersect in O. Join CO, FO. It is required to prove that CO,
FO are in one straight line.
Dem.—The L ABO =* DEO (xxix.), and the L AOB = DOE
(xv.), and the side AB = DE (hyp.); .*. (xxyi.) BO *= 3B0.
Again (xxix.) the L CBO = FEO, and CB = EF (hyp.), and we
have shown that BO = EO;	(iv.) the L BOC = EOF; to
each add the L FOB, and we have BOC + FOB = EOF + FOB;
but EOF + FOB = two right angles (xm.); .·. BOC + FOB *=
two right angles, and .·. (xiv.) CO, OF are in one straight
line.
PROPOSITION XXXI.
1. Let A, B, be the given L *, and H the altitude.
Sol.—Draw any line CD, and make the L DCE = A, and the
L CDE ss B; let fall a X EF on CD. If EF = H, the Δ is
constructed. If not, produce it) and cut off EG — H. Through
G draw JK’ J to CD, and produce EC, ED, to meet-it inBOOK I.]
EXERCISES ON EUCLID.
19
Dem.—The L EJK = ECD (xxix.) = A. In like manner
EKJ = B, and EG = H. Therefore EJK is the Δ required.
2.	Let AB be the given line, C the given point, and M the
given L *
Sol.—Through C draw CE || to AB (xxx.). . At the point C in
CEmake the L ECD = M. The L ECD = CDA (xxix.)
:·. CDA = M.
3.	Dem.—The L CAD = ADE (xxix.); but CAD = EAD
•(const.); ADE = EAD, and .·. EA - ED. In like manner
FB = FD. Again, the L CAB = DEF (xxrx.); hut CAB is an
L of an equilateral Δ; DEF is an L of an equilateral Λ.
Similarly DFE is an L of an equilateral Δ; hence DEF ti an
equilateral Δ; .·. DE — EF; but DE = AE; AE = EF* In
like manner BF = EF. Hence AB is trisected,
4.	Let ABC be the equilateral triangle.
Sol.—Let fall a JL AD on BC. Bisect the L BAD .by AE,
meeting BC in E. Through E draw EF || to AD, meeting AB
in F. Through F draw FG j| to BC, and complete the □ EFGH.
EFGH is a square.
Dem.·—The L FEA = EAD (xxix.), = FAE; .·. FA = FE ;
but FAG is an equilateral Δ, because FG is || to BC; .·. AF
= FG; but AF = EF; .·. EF = GF, and EF = GH, and GF
= EH; .·. the four sides are equal, and (xxix.) the L GFE
= BEF; but BEF is a right L, GFE is right. Hence EFGH
is a square.
δ. (1) Let ABC be the triangle.
Sol.—Produce AB to G. Bisect the L GBC by BF, meeting AC
produced in F. Through F draw FG || to BC.
Dem.—The L CBF = BFG (xxix); but CBF = GBF (const.);
.·. GBF = BFG, and .·. FG = BG. If we bisect the L BCF, we
get another solution.
(2)	Sol.—Produce AB, AC to E, F. Bisect the L* CBE,
BCF ; and through D, where the bisectors meet, draw EF (| to
BC, meeting AE, AF, in E, F.
Dem.—The L CBD = EDB (xxix.); but CBD = EBD*(const.);
EDB = EBD ; and (vi.) EB s= ED. Similarly, FC « FI).
Hence EB + FC = EF. ;
If, we bisect the Le ABC, ACB, we have another solution.
(3)	Sol.-^Produce the base BC to G. Bisect the L% ABC,
ACG, by BD, CD. Through D draw DF || to BC,'-meeting AB,
AC in F, E.20
EXERCISES ON EUCLID.
[book I.
Dem.—The L FDR = CBD (xxix.); but CBD «= FBD*
(const.); .·. FBD = FDB; and therefore FB = FD. Jn like-
manner CE = DE. Hence BF — CE = FD — DE = FE. If we
produce CB to H, and bisect the Z.8 ACB, ABH, we will have
another solution.
6.	Let AB, DC be the || lines, and B, D the given points.
Sol.—Join BD; bisect it in E. At E erect EA J, to BD V
produce it to meet CD in C. Join AD, BC.
Dem.—Because EB = ED, EA common, and the L AEB
= AED; .·. (iv.) AB = AD. In like manner BC = DC, and the
four sides are equal to each other. Hence (Def · xxix.) ABCD is-
a lozenge.
7.	Let AB, AC be the lines given in position, M the line of
given length, and FG the line to which the required line is to be
parallel.
Sol.—(1) In FG take a part DE = M; through D draw D^ j}
to AC, and through B draw BC || to DE. BC fulfils the requi*
conditions.
v	M
Dem.—Because DBCE is a parallelogram, BC = DE; but DE
= M; i\ BC = M.
(2) Sol.—In FG take FH = M. Through F draw FJJ( to
AB, meeting CA produced in J; and through J draw JK
| to FH. JK fulfils the required conditions.
Dem.—Because FJKH is a parallelogram, FH = JK; but
FHa.M; .·. JK*=M.©00X I.]
EXERCISES ON EUCLID»
21
PROPOSITION XXXII.
1.	Let ABC be the right angle.
Sol.—Make the L ABD equal an L of an equilateral Δ (xxm.)9
and draw BE bisecting it.
Dem.—Because the L ABD is an angle of an equilateral Δ9 it
da two-thirds of a right L; CDB is one-third, and half ABD is
one-third. Hence ABC is trisected.
2.	(1) Let ABC be the triangle.
Dem.—Draw the median AD. Now if BD be greater than
AD, the L BAD will be greater than ABD (xvm.) Similarly the
L CAD will be greater than ACD. Hence the L BAC will be
greater than ABC + BCA, and .·. will be obtuse, when the side
BC is greater than 2AD.
(2)	Dem.—If BD = AD, the L BAD = ABD; and if CB = AD,
the L CAD = ACD. Hence the L BAC is = ABC + BCA, and
-.·. right when BC = 2AD.
(3)	In like manner it can be shown that the L BAC is acute,
when BC is less than 2AD.
3.	Let ABCDE be the polygon.
Dem.—Produce AB, DC to meet in A'; BC, ED to meet
in B', &c.
Now the sum of the L* of the Δ BA'C is two right A·;
similarly the sum of the L8 of each of the external Δ* is two right
L *. Hence if there be n external triangles, the sum of their L8
will be 2n right L8; but the sum of the exterior L ■ BCA',
CDB', &c., is four right L8; and the sum of the exterior L8
CBA', DCB', &c., is four right L\ Hence the sum of the
remaining L8 must be (2n - 8) right L8; that is, 2(n — 4) right Ls.
4.	Let BAC be the triangle.
Dem.—Produce BA to D, and bisect the L CAD by the line
AE || to BC.
The L EAC = ACB (xxix.); but EAC = EAD, and EAD
= ABC; .·. ACB = ABC. And hence AB = AC.
5.	Let E be the point where CD cuts AB.
Dem.—Bisect AB in F. Join CF, DF. Now the lines AF,
BF, CF, DF are equal (xn., Ex. 2). And because FD = FB, the
L FBD = FDB = FDE + EDB; to each* add the L EDB;
^hen the L8 EBD + EDB « FDE + 2EDB; but the L CEB
= EBD + EDB (xxxn.); .·. CEB = FDE + 2EDB; but CEB
^FCB + CFE, and FCD = FDE; .·. CFE = 2EDB. Again,22	EXERCISES ON EUCLID.	[book i.
CFE = ACF + CAF; but ACF = CAF (v.); .·. CFE = 2CAF,
. ·. 2CAF = 2EDB. And hence CAF = EDB.
6.	Let ABC be the triangle.
From B, C draw ±8 BD, CE to the sides AC, AB, and let them
meet in G; join AG, and produce it to meet BC in F. It
is required to prove that AF is ± to BC.
Bern.—Join DE. Now we have two right-angled triangle®
BEC, BDC, and we have joined their vertices E, D; hence (6)
the L EDB = ECB. Similarly from the Δ· AEG, ADG, the L
EAG = EDG (5); .·. EAG = FCG, and AGE = CGF (xv.);
hence (Cor. 2) the L AEG = GFC; but AEG is a right angle ;
CFG is right; and hence AF is _L to BC.
7.	Let ABCD be the □, and· BE, CE the bisectors of tho
adjacent angles B, C. It is required to prove that the L BEC
is right.
Dem.—The L 8 ABC, DCB equal two right angles (xxix);
EBC + ECB equal a right angle); and hence the L BEC is right.
8.	Let ABCD be the quadrilateral. Bisect the external L8 A,
B, C, D; let the bisectors meet in E, F, G, H. It is required
to prove that the L8 EHG, EFG, of the quadrilateral EFGH,
are together equal to two right angles.
Dem.— Produce BA, CD to J, K. Now the L8 ADC, ADK,
DAB, DAJ equal four right angles; and the L8 DHA, HAD,
ADH equal two right angles; .·. the Z.8 of the Δ HAD equal
half sum of the L8 ADC, ADK, DAB, DAJ; but the L8 HAD,
ADH are the halves of JAD, ADK; hence the L DHA is half
sum of BAD, ADC; in like manner BFC is half sum of ABC,
BCD. Hence the sum of the angles DHA, BFC is half sum of tho
four angles of the quadrilateral ABCD, and equal to two right
angles.
9.	Let the sides of the triangle DEF be perpendicular to th&
sides of ABC. It is required to prove that the Δ8 DEF, AB$
are equiangular.BOOK t.]
EXERCISES ON EUCLID.
28
• Dem.—Since the L · CHE, EGC are right, the sum of the L *
HCG + HEG = two right L · (Cor. 3), and HED + HEG =5 two
right l *. Reject the common L HEG, and we have the L HCG
= DEF, that is, the L ACE = DEF. In like manner the. L
BAG = EFD, and ABC = EDF.
10. (1) Let M equal sum of sides, and N the hypotenuse.. .
Sol.—Draw any line AC, and make it equal to M. In AC take
a part AD = N. At the point C in AC make the L ACB equal
half a right angle. With A as centre, and AD as radius, describe
the Θ DBF, cutting CB in B. Join AB, and at the point B in
BC make the L EBC = ACB. AEB is the required triangle.
Dem.—Because the L EBC = ACB, EC =EB (vi.). To each
add AE, and we have AC = AE + EB; but AC = M (const.);
.·. AE + EB = M. Again, the L AEB = EBC + ECB (xxxii.) ;
but EBC = ECB; AEB = 2ECB, and is therefore a right
. (2) Let M equal differences of sides, and N the hypotenuse.
Sol.—Draw any line AC = N. In AC take AD = M. At the
point D in AC make the L CDB = half a right angle. With A
as centre, and AC as radius, describe the Θ CBF, cutting DB in
B. From B let fall the _L BE on AC. Join AB. AEB is the
required triangle.
I>em.—Because the L AEB is right, and EDB half right,
.·. EBD is half right, and (vi.) ED = EB. Hence AD is the24
EXERCISES ON EUCLID,
[books»
difference between AE and EB. Again, AC = AB; but AC = N.
Hence AB = N.
11» Let ABC be an isosceles A. From B let fall a X BD on
AC. It is required to prove that the L DBC = half BAC.
Bern.—Bisect the L BAC by AE, meeting BC in E, and BD
in F.. Now the L BFE = AFD (xv.), and BEF = ADF
(ix., Ex. 2); hence (Cor. 2) EBF = FAD; but FAD = half
BAC. Therefore EBF = half BAC.
12.	Let ABC be a triangle. Produce BC to E. Bisect the
L· ABC, ACE by BD, CD. It is required to prove that the
L BDC = half BAC.
Bern.—The L ACE = ABC + BAC (xxxii.), and DCE = DBC
+ BDC; but DCE = £ ACE; .-. DBC + BDC = £(ABC + BAC);
but DBC = £ ABC. Hence BDC = £ BAC.
13.	Bern.—Because AB = AD, the L ADB = ABD; but
ADB = ACB + CBD (xxxii.) ; .·. ABD = ACB + CBD. To
each add the L CBD, and we have the L ABC = ACB + 2CBD;
2 CBD = ABC - ACB ; and hence CBD = \ (ABC - ACB).
14.	Bern.—Produce BA, BC to D, Έ. Bisect the L * CAD,
ACE by AF, CF.
Now the L ACE = (A + B) (xxxii.) ; .·. ACF = i (A + B).
Similarly the L CAF = J (B + C). Hence the L AFC
» £(C + A).
1. Bern.—Join AD, BE, CF. Now, because AB is equal and
parallel to DE, (xxxiii.) AD is equal and parallel to BE.
In like manner CF is equal and parallel to BE; hence CF is
equal and parallel to AD; and ·*. (xxxiii.) AC is equal and
parallel to DF.
2. (1) Let AB, CD be equal and parallel lines, and EH any
other line. From A, B, C, D let fall Xs AE, BG> CF, DH on
EH. It is required to prove that EG = FH.
PROPOSITION XXXIII.
B
D F26
BOOK i.]	EXERCISES OK EUCLID.
Dem.—Through A, G draw AJ, CK || to EF.
E F
G H
Now» because AJ, CK are each || to EF, they are || to one
another, and AB is [J to CD; hence (xxix., Ex. 8) the L BAJ
=3 DCK, also the L AJB = CKD, because each is right, and
the side AB = CD ; .·. (xxvj.) AJ = CK; but AJ = EG,
and CK = FH. Hence EG = FH.
(2) As in (1) the L* BAJ, AJB are respectively equal to the
L 8 DCK, CKD, and the side AJ = CK. Hence AB - CD.
3.	Dem.—Since AB=CD, and AJ = CK, and the L AJB=CKD,
each being right; .·. (xxvi., Ex. 6) the A8 ABJ, CDK are equal
in every respect; hence the L ABJ = CDK; but CDK = CMG
(xxix.); .*. ABJ = CMG. Hence AB is parallel to CD.
4.	Let AB, CD be the equal and parallel lines. Join AD, BC,
intersecting in E. It is required to prove that AD, BC bisect
each other in E.
Dem.—The L8 ABE, BAE are respectivly equal to the
L· DCE, EDC, and the side AB = CD (hyp.). Hence (xxvi.)
BE — CE, and AE = DE.
PBOPOSITION XXXIY.
1.	See last exercise to Prop, xxxiii.
2.	Let ABCD be the O; AC, BD its diagonals, which are
equal. It is required to prove that the L* of ABCD are
right LK
Dem.—Because AD = BC, and AB common, and the bases
BD, AC equal, (vm.) the L BAD = ABC; but (xxix.) BAD
+ ABC equal two right angles; hence each is right, and (xxxiv.)
the L BAD = BCD, and ABC = ADC. Therefore all the L*
are right angles.
3.	See “ Sequel to Euclid,” Prop, xv., p. 11, 3rd Edition.EXERCISES ON EUCLID,
[BOOK I,
4.	Let AB, CD be two H lines, of which AB is the greater.
Join AC, BD. It is required to prove that AC, BD produced
will meet.
Dem.—From BA cut off EB = CD. Join EC. Because EB
is equal and parallel to CD; (xxxiii.)EC is equal and parallel
to BD; and .*. (xxix.) the L AEC = ABD. To each add the
L CAE; then CAE + AEC = CAE + ABD; but CAE and AEC
are less than two right angles (xvii.) ; hence CAE and ABD are
less than two right angles. And. *. AC, BD, if produced, will meet.
5.	Let ABCD be a quadrilateral, having AB, CD parallel, but
not equal; and AC, BD equal, but not parallel. It is required to
prove that the L8 CAB, CBD are supplemental.
Dem.—In CD take CE = AB. Join BE. Now (xxxm.)
AC is = and || to BE; but AC = BD*(hyp.); .·. BE = BD;
and .·. (v.) the L BDE = BED, and (xxxiv.) the L CAB
= CEB; hence the L8 CAB -f BDE = CEB + BED. But CEB
and BED are supplemental; hence CAB and BDE are sup-
plemental.
6.	Let A, B, C be the middle points of the sides.
Sol.—Join AB, BC, CA; and through the points A, B, C draw
DE, EF, FD || to BC, AC, AB. DEF is the required triangle.
Dem.—AB = CD (xxxiv.), and AB = CF; hence CD = CF.
In like manner AD = AE, and BF = BE.
7.	Let ABCD be a quadrilateral, whose diagonals are AC, BD.
Through B, D, draw FG, EH || to AC, and through C, A, draw
GH, EF || to BD·. Join FH. It is required to prove that the
area of the Δ EFH is equal to the area of ABCD.
Dem.—The area of the Δ EFH is half the area of the □ EFGH
(xxxiv.), and the area of ABCD is half the area of EFGH;
.*. EFH = ABCD ; -and the sides EF, EH are equal to BD, AC ;
and the L FEH = AJD, which is the L between AC, BD.
PROPOSITION XXXYI.
Dem.—Produce AB, EF to meet in J. Through.J draw
JK || to AH or BG, and produce DC to meet it in K. Join KG.
Now JK = BC (xxxiv.); but BC = FG (hyp.); .·. JK = FG,
and it is || to it; hence JFGK is a □; .·. JF is || to KG; but
JE is || to GH. Hence KG, GH are in one straight line; .*. JEHK
is a □ and it is equal to JADK (xxxv.); but J.BCK = JFGK.,
Hence ABCD = EFGH.BOOK I.]
EXERCISES ON EUCLID.
27
PROPOSITION XXXVII.
1.	See “ Sequel to Euclid,” Prop, vi., p. 4, 3rd Edition.
2.	Let ABCD be a given quadrilateral. It is required to con-
struct a triangle equal in area to ABCD.
Sol.—Join AC. Produce DC to E ; and through. B draw BE
|| to AC. Join AE. ADE is the Δ required.
Dem.—The Δ* ABC, AEC are equal (xxxvii.) To each add
the Δ ACD, and we have the Δ ADE equal to the quadri-
lateral ABCD.
3.	Let the pentagon ABCDE be the given rectilineal figure. It
is required to construct a Δ equal in area to ABCDE.
Sol.—Join AC, AD. Through B, E draw BF, EG |j to AC,
AD, and meeting DC produced both ways in F, G. Join AF,
AG. AGF is the Δ required.
Dem.—The Δ8 ABC, AFC are equal (xxxvn.); to each add
ACDE, and we have the pentagon ABCDE equal to the quadri-
lateral AFDE. Again (xxxvn.), the Δ AGD = AED. To each
add the Δ ADF, and we have the Δ AGF equal to the quadri-
lateral AFDE; but AFDE = ABCDE. Hence AGF = ABCDE.
4.	Let ABCD be a given parallelogram. It is required to con-
struct a lozenge equal to ABCD, and having CD as base.
Sol.-—If AD = DC, the thing required is done. If not, let DO
be the greater. With D as centre, and DC as radius, describe
a Θ ECG, cutting AB in E. Join DE. Through C draw CF
|| to DE, meeting AB produced in F. DEFC is the required lozenge.
Dem.—DE = DC; but DC = EF (xxxiv.); .·. DE = EF.
Hence the four sides are equal; .·. DEFC is a lozenge, and
(xxxv.) is equal to ABCD.
5.	Let ABC be a Δ, whose base BC is given, and whose area
is given. It is required to find the locus of its vertex A.
Sol.—Through A draw DE || to BC. DE is the required locus.
Dem.—Take any other point F in DE. Join BF, CF. Now
(xxxvn.) the Δ8 ARC, FBC are equal. Hence DE is the locus
of the vertex of all triangles, having BC as base, and whose area
is equal to the area of the Δ ABC.
6.	Dfem.—Through E draw EG || to FD, and meeting AD
in G. Join GF, GC. Now (xxxvii.) the Δ EFD = GFD; but
GFD =‘GCD, and GCD is less than ACD; .*. EFD is less than
ACD, that is, is less than half ABCD.28
EXERCISES ON EUCLID.
[rook I.
PROPOSITION XXXVIII.
1.	Let ARC be the Δ, and AD one of its medians. It is
required to prove that AD bisects the triangle.
Dem.—BD =■ CD (Def. Prop, xx.); .·. (xxxvm.) the Δ ABD
= ACD.
2.	Let ABC, DEF he two ΔΒ, having the sides AB, BC equal
o the sides DE, EF, and the contained L8 supplemental. It is
required to prove that the Δ* are equal.
Dem.—Produce CB to G, and make BG = BC or EF. Join
AG. Now the L8 ABC, DEF are supplements (hyp.), and ABC,
ABG are supplements (xm.) Reject ABC, and we have ABG
= DEF; hence (iv.) the Δ ABG « DEF; but ABG = ABC
xxxvm.) Hence DEF = ABC.
3.	Dem.—Divide the base BC of the Δ ABC into any number,
such as four equal parts, in the points D, E, F. Join AD, AE,
AF. It is required to prove that the four Δ8 into which ABC is
divided are equal.
The Δ BAD = EAD (xxxvm.) Similarly EAD = EAF, and
EAF = CAF. Hence the four Δ8 are equal.
4.	Let ABDC he a □, whose diagonals AD, BC intersect in F.
In BC take a point E. Join EA, ED. It is required to prove
that the Δ ABE = DBE, and that ACE = DCE.
Dem.—AF = DF (xxxiv., Ex. 1); hence (xxxvm.) the Δ AFB
= DFB, and AFE = DFE; hence AEB = DEB; hut ABC = DBC;
.·. AEC = DEC.
5.	Let ABCD be a quadrilateral; and let AC, one of its diago-
nals, bisect the other, BD in E. It is required to prove that AC
bisects ABCD.
Dem.—The Δ AEB * AED (xxxvm.), and the Δ CEB = CED.
Hence ABC = ADC.
6.	See “ Sequel to Euclid,” Prop, xm., p. 10, 3rd Edition.
7.	See “ Sequel to Euclid,” Prop, xm., p. 10, Cor. 1.
8.	See “ Sequel to Euclid,” Prop m., Cor 1, p. 2.
9.	Let ABC be a Δ ; D, E the middle points of AB, AC;
F any point in BC. Join DE, EF, FD. It is required to prove
that DEF = J ABC.
Dem.—Bisect BC in G. Join DG, EG. Now (xxxvii.) the
ΔDEF = DEG; hut DEG = \ ABC (8). Hence DEF = \ ABC.
10.	Let ABC be a given Δ, and D a given point in BC. It is
required to draw a line through D, bisecting the Δ ABC.BOOS I.]
EXERCISES ON EUCLID.
Sol.—Join AD. Bisect BC in E. Through E draw EF || to*
AD, and meeting AB in F. Join DF. DF is the required line.
Dem.—Join AE. Now (xxxvn.) the Δ* EFD, EFA are equal.
To each add the Δ BEF, and we have the Δ BFD = BAE; hut BAE
= JBAa Hence BFD = 4 BAC.
11.	Let ABC be a given Δ, and D a given point within it. It
is required to trisect ABC by three lines drawn from D.
Sol.—Trisect BC in E, F (xxxiv., Ex. 3) Join AD, DE, DF.
Through A draw AG, AH || to DE, DF. JoinDG, DH. AD,.
DG, DH trisect ABC.
Dem.—Join AE, AF. Now (xxxvii.) the Δ* ADG, AEG are
equal. To each add the Δ AGB, and we have the quadrilateral
ADGB equal to the Δ AEB; hut AEB = £ ABC (3); hence
ADGB = J ABC. In like manner ADHC = £ ABC; the
ADGH = £ ABC. Hence thejA ABC is trisected by the lines
AD, GD, HD.
12.	Let ABCD be a □, whose diagonals AC, BD intersect
in E. Through E draw any line FG, meeting AB, CD in F, G.
It is required to prove that FG bisects ABCD.
Dem.—The L BEF = GED (xv.), and the L FBE = GDE
(xxix.), and the side EB = ED (xxxiv., Ex. 1); hence (xxvi.)
the Δ8 BEF, DEG are equal. Similarly, AEF = CEG, and
AED = CEB. Hence FG bisects ABCD.
13.	Let ABCD be a trapezium. Bisect AD in E. Join EBr
EC. It is required to prove that the A BEC = J ABCD.
Dem.—Produce BE, CD to meet in F. Now (xxvi.) the
A AEB = DEF, and EB = EF. And since AEB = DEF,
AEB + CED = CEF; but (xxxvih.) CEF = BEC. Hence
BEC = AEB + CED.
PROPOSITION XL.
1. Let ABC, DEF be two A8 whose bases and altitudes are
equal. It is required to prove that the A8 are equal.
Dem.—Produce BC; and in BC produced cut off GH = EF
or BC, and construct the A JGH, having its sides JG, GH, HJ
respectively equal to the sides DE, EF, FD of the A DEF.
Join AJ; and from A, J let fall ±8 AL, JK on BH. Because
the A DEF = JGH, their altitudes are equal; but the altitudes
of DEF and ABC are equal (hyp.); hence the altitudes of JGH-$0
EXERCISES ON EUCLID.
£book i*
and ABC are equal; that is, JK = AL, and they are parallel;
hence (xxxm.) AJ, BH are parallel; *·. (xxxvm.) the Δ ABC
= JGH ; hut JGH = DEF. Hence ABC = DEF. *
3.	See	“ Sequel to Euclid,”	Prop, n., p. 2.
4.	See	“Sequel to Euclid,”	Prop, m., Cor.	1,	p. 2.
h. See	“ Sequel to Euclid,”	Prop, ii., Cor.,	p.	2.
6.	See	“ Sequel to Euclid,”	Prop, y., p. 3.
7.	Let ABCD he a trapezium, whose opposite sides AD, BC
are ||; E, F the middle points of AB, DC. Join EF. It is
required to prove that AD + BC = 2EF. .
Dem.—Through A draw AH || to' DC, meeting EF, BC
in G, H.
Now (xxxiv.) AD = GF, and HC = GF; . ■*. AD + HC « 2GF,
and (5) BH = 2EG. Hence AD + BC = 2EF.
8.	See “ Sequel to Euclid,” Prop, m., Cor. 2, p. 3.
9.	Let ABCD he a quadrilateral; AC, BD its diagonals. Bisect
AC, BD in E, F. JoinEF. Bisect AB, CD, BC, AD in G, H, J, K.
Join GH, JK. It is required to prove that the lines EF, GH, JK
are concurrent. .
Dem.—Join EG, EH, FG, FH, GJ, GK, HJ, HK.
B
Now ((2) and (5)) GF is || to AD, and =± J AD. Similarly, EH is
J1 to AD, and = J AD; hence GF is = and || to EH ; .·. (ixxra.)
GFHE is a □; hence (xxxrv.,1) the diagonal EF bisects GH
in L. In like manner GJHK is a O, and the diagonal JK
bisects GH. Hfence the lines EF, GH, JK are concurrent.■BOOK I.]
EXERCISES OK EUCLID.
31
PROPOSITION XLV.
1.	Let A and B be two rectilineal figures. It is required to
«construct a rectangle equal to the sum of A and B.
Sol.—Construct a rectangular parallelogram EFGH equal to A
■(xlv.), and to the straight line GH apply a □ GHIK equal to B,
and haying the L GHI a right angle. FI is the required reet-
,ungle.
Bern.—The figure FI is equal to the sum of A and B, and it
is evidently a rectangle.
2.	If we apply the □ GHIK to the left of GH, it is evident
, that EFKI 'will be the required rectangle.
PROPOSITION XLYI.
1.	(1) Let AB, CD be equal lines. Upon AB, CD describe
squares ABEF, CDGH. It is required to prove that ABEF
= CDGH.
Bern.—Join AE, CG. Now AB = BE, and CD = DG; but
AB = CD ; hence AB and BE «= CD and DG, and the L ABE
= CDG; .·. (iv.) the Δ ABE = CDG; but ABEF = 2ABE,
and CDGH = 2CDG. Hence ABEF = CDGH.
(2) Let ABEF » CDGH. It is required to prove that
AB = CD.
Bern.—If not, from AB cut off AJ = CD ; and on AJ describe
... the square AJKL. Now since AJ = CD, AJKL = CDGH; but
. CDGH = ABEF (hyp.); ./. AJKL = ABEF, which is absurd.
Hence AB = CD.
2.	Let ABCD be a square, and BD one of its diagonals. In
3D take a point E, and through E draw FG, HJ j| to AB, AD.
It is required to prove that HG, FJ are squares.
Bern.—The L ADB = ABD (v.); but ADB = HEB (xxix.);
;·. ABD= HEB ; hence the side HE = HB ; but HB = EG, and
HE = BG; .*. HB, HE, GB, EG are all equal. Again, the
L% EHB, GBH equal two right L9: but GBH is right; .*. EHB
is right, and (xxxiv.) the opposite /.“are equal. Hence EGBH
is a square. In like manner EJDF is a square.
3.	Let ABCD be a square, and E, F, G, H points in the sides
. AB, BC, CD, DA reepectively equidistant from A, B, C, D. Join
JEF, FG, GH, H J. It is required to prove that EFGH is a square.32
EXERCISES ON EUCLID.
[book I»
Dem.—The A· AHE, BEF are equal in every respect (nr.);
the side EH = EF. Similariy, EF = GF, and EH = GH.
Hence the four sides are equal. Again, the L AHE = BEF»
To each add the L ΑΈΗ, and we have the L* AHE, AEH equal
to the L * BEF, AEH; hut AHE + AEH = a right L, since the L
at A is right; BEF + AEH = a right L. Hence the L
FEH is right. In like manner the other L* are right;
EFGH is a square.
4. Let ABCD be a square. It is required to divide it into five
equal parts, namely, four right-angled triangles and a square.
Sol.—Divide AC into five equal parts, and let AE = $ AC*
Through E draw EF || to AB. Upon AB describe the semicircle
AGHB, cutting EF in the points G, H. Join AG, and produce
it. From C let fall a JL CK on AK, and produce it. Join BG.
.From D let fall DM _L to BG, meeting CK produced in L.
XBCD is divided into five equal parts.
Dem.—Join OG. Because O is the centre of AGHB, OG = OA;
.·. (v.) the L OAG=OGA. Similarly, the L OBG=OGB. Hence
(xxxn., Cor. 7) the L AGB is right. Again, since the L AKCis
right, the L* KCA, KAC are together equal to a rights, and
therefore equal to the L CAB, which is right. Reject the L
KAC, and we have the L KCA = KAB, and the L CKA = AGB,
because each is right, and the side AC = AB; hence (xxyi.) the
AAKC = AGB; AK = BG, and CK =* AG. In like manner it
can be shown that the A* CLD, BMD are each equal to AGB.
Hence the four A* are equal, and the lines AK, BG, CL, DM
are equal, and also the lines AG, BM, CK, DL; hence the re*
mainders GK, GM, LK, LM, are equal. Again, the rectangle
ABEF is } ABCD, and the A AGB is } ABEF; .-. AGB is £
ABCD; .·. AKC, CLD, BMD are each £ ABCD. Hence KGML
must be i ABCD, and it is a square, for we have proved the rides
•equal, and the L* are right angle·.EXERCISES ON EUCLID.
PROPOSITION XLYII.
1.	Dem.—ACHK=AOLG; bat AOLGis therectangle AG. AO;
that is, AB . AO, and ACHK is AC3. Hence AC* = AB . AO.
Similarly, BC3=AB . BO.
2.	Dem.—From GA cut off GM = GL, and draw MN fl to GL.
Now the figure AL = AH (xlvii. ); but AH = AC* = AO* + OC*;
and GN = MN* = AO*; hence OM = CO*; but OM = AO. OB,
since ON = OB. Hence CO* = AO. OB.
3.	Dem.—AC* = AO* + OC*, and BC* = BO2 + OC*. Sub-
tracting, we get AC* — BC* - AO2 — BO*.
4.	Let AB, CD be the lines whose squares are given. It is
required to find a line whose square shall be equal to the sum of
the squares on AB and CD.
Sol.—Erect AE _L to AB, and make it equal to CD. Join BE.
Now (xlvii.) BE* = AB* 4 AE* = AB* 4 CD*.
5.	Let ACB be a A whose base AB is given, and the difference
of the squares of its sides. It is required to prove that the locus
of C is a right line X to AB.
Dem.—From C let fall a X CO on AB. Now (3) AC* — BC*
= AO* - BO*; but AC* - BC* is given; .·. AO* — BO* is given,
and .*. 0 is a given point; .*. the line OC is given in position.
Hence OC is the locus of C.
6.	Dem.—Let P, Q be the points in which AC, GC intersect
BK. Now (rv.) the A1 CAG, BAK are equal in every respect;
.·. the L ACG=sAKB, and the L CPQ = APK (xv.); .*. (xxxii.,
Cor. 7) the L CQP = KAP; .·. CQP is a right L, and CG is X
toBK.
7.	See “ Sequel to Euclid,” Book I., Prop, xxiii. (3).
8.	Dem.—Since EB = AH, AB = AE + AH, and AC is the
square on AB; .*. AC is equal to the square on the sum of AE
and AH; but AC exceeds EG by four times the Δ AEH, and EG
is the square on EH; hence the square on the sum of AE and AH
exceeds the square on EH by four times the A AEH.
9.	Dem.—Join PH, QC. Now (xxxvn.) the A PCQ = PBQ.
To each add APQ, and we have the A ACQ = APB. Again, the
sum of the A* KAP, HCP equals } KC, and the Δ KAB = J KC
(xu.); .·. KAB = KAP and HCP. Reject the A KAP, and we have
the A APB = HCP; but APB = AQC; hence HCP = AQC, and
their bases HC, AC are equal. Hence (xl.) their altitudes PQ,
PC are equal.
D34
EXERCISES ON EUCLID.
[book. I*
10.	J See “ Sequel to Euclid,” Book I., Prop. xxm. (2).
11.	Let Μ, N be two lines. It is required to find a line whose
square shall be equal to Μ2 — N2.
Sol.—Draw a line AB = M, and in it take AC = N. Erect
CE X to AB. With A as centre, and AB as radius, describe
a Θ cutting CE in D. CD is the required line.
Dem.—Join AD. Now AD2 = AC2 + CD2; .*. CD2 = AD2
- AC2 = AB2 - AC2 = M2- N2.
12.	Dem.—From AC cut off AD = BC; then, evidently, CD is
the difference between AC and CB. On AB describe a square
ABFG, and on CD describe a square CDEH; and produce DE,
EH to. meet ABFG in G, F (figure similar to that on p.. 89,
<4 Elements ”).
Now ΊΕ is less than AF by the sum of the four triangles; that
is, by four times the Δ ABC. Hence CD2 -f 4 ABC = AB2.
13.	Dem.—Join CF, CG, cutting AE, BE in P, Q. Through
A draw AM || to GC, cutting BE in R, and meeting LC produced
in M. Join BM, cutting AE in N. Now, because AM is y to
GC, and AG to ML, AGCM is a D; .·. AG = CM; but AG = BF;
.*. BF = CM; .*. FCMB is aO; .*. CF is || to BM; hence
(xxix.) the L ANM = APC; but APC is a right L (6); .*. ANM
.is right, and AN is X to BM. In like manner BR is X to AM;
and OM being X to AB, .·. AN, BR, OM are the Xs of the
Δ AMB ; .*. (xxxii., Ex. 6) these lines are concurrent; that is,
the lines AE, BE, CL are concurrent.
14.	Let ABC be an equilateral triangle. Let fall a X AD on BC.
Dem.—AB2 = AD2 + BD2 (xlvii.) ; .·. 4 AB2 = 4 AD* +4 BD2,
but AB2 = 4 BD2, since AB = BC — 2 BD. Subtracting, we get
3 AB2 = 4 AD2.
15.	Sol.—In AB take AH = BG. Join DH, FH. These lines
divide the figure into the parts required.fftOOK Γ.]
EXERCISES ON'EUCLID.
35
Dem.—For if we take the Δ AHD and place it in the position
DCK, and place the A FHG in the position FEE, the figure
HFKD will he equal to the figure AGFECD; and it is evidently
4* square.
16.	Let AB be the hypotenuse of the right-angled A ACB.
Bisect BC, AC in D, E. Join AD» BE. It is required to prove
•that 4 AD2 + 4 BE2 = 6 AB2,
Dem.—4AD2 = 4AC2+ 4 CD2; but BC^ 4 CD* ; .·. 4 AD2
-= 4 AC2 4* BC2. Similarly, 4 BE2 = 4 BC2 + AC2. Adding* we
get 4 (AD* + BE2) = 5 (AC2 + BC2) = 5 AB2.
17.	Let ABC be a Δ, and Oa point within it. Through 0
-draw J_8 AD, BE, CF to BC, CA, AB. It is required to prove
that AF2 + BD2 + CE2 = BF2 + DC2 4- EA2. Now (2) AF* - BF2
= AO2 - BO*; BD2 - CD2 = BO2 - CO2; and CE2- AE2= CO2
- OA2. Adding, we get AF2 + BD2 + CE2 - (BF2 + DC2 + EA2)
= 0; and hence AF2 + BD2 -f CE2 = BF2 4- DC2 + EA2. Simi-
larly for a figure of any number of sides.
18.	Let ABCD be a rectangle, and 0 any point. Join OA, OB,
*OC, OD. It ie required to prove that OA2 + OC2 = OB2 + OD2.
Dem.—Produce DA, CB to F, G, and let fall _L8 OF, OG
<m DF, CG.
Now, OD2 =* DF2 + OF2; and OA2 = AF2 4 OF2; .·. OD2
- OA2 = DF2 - AF2. Similarly, OC2 - OB2 = CG2 - GB2 ; but
DF2 = CG2, and AF2 = GB2; .·. OD2 - OA2 = OC2 - OB2; and,
by transposition, we have OD2 4- OB2 = OC2 + OA2.
19.	Let AB be the hypotenuse of a right-angled Δ ABC. It
is required to divide it into two parts, such that the difference of
their squares shall equal AC2.
Sol.—Bisect BC in D. Join AD, and let fall the ± DE on
AB. AE2 - BE2 = AC2.
Dem.—AD2 - BD2 = AE2- BE2 (3); that is, AC2 + CD2 - BD2
= AE* - BE2; but CD2 r= BD2 (const.); .·. AC2 = AE2 - BE2.
d236
EXEBCJSES ON EUCLID.
[book k
20. Let ABC be the Δ. From B, C let fall X8 BE, CD on
AC, AB. It is required to prove that AB. BD -f- AC.CE * BC2.
Dem.—On BC describe a square BCFG. Produce BE) CD to
H, J; and through B, C draw BL, CK || to DJ, EH, and mako
BL = AB, and CK = AC. Complete the □· BLJD, CKHE.
Draw AM || to CF, meeting GF in M. Now it can be shown, as
in (xlvii.), that BM = BJ, and CM = CH; BF = BJ + CH;
but BF - BC2, BJ = AB. BD, and CH = AC. CE. Hence BC21
= AB.BD +AC.CE.
Miscellaneous Exercises on Book I.
1.	See “ Sequel to Euclid,” Book I., Prop, hi., Cor. 1.
2.	Let DEF be the original Δ, ABC the Δ formed by drawing'
through each vertex a || to the opposite side. Let fall a X FG on
DE. It is required to prove that GF bisects BC perpendicularly.
Dem.—The L CFG = DGF (xxix.); but DGF is right; .·. CFG
is right. Again, BF = DE (xxxiv.), and CF = DE; .*. BF = CF.
Hence GF bisects BC perpendicularly. Similarly, the Xs from.
D, E on EF, DF bisect AB, AC perpendicularly.
3.	Let ABC be a given L, and D a given point. It is required;
to draw a line through Ϊ), so that the parts DA, DC, intercepted
by AB, BC, may be equal.
Sol.—Through D draw DE || to AB, meeting BC in E, and
make EC = BE. Join CD, and produce it to meet AB in A.
Dem.*—AC is bisected in D (xl., Ex. 3).
4.	Let BD, CE, two of the medians of the Δ ABC, intersect
in H. Join AH, and produce it to meet BC in F. It is required
to prove that AF is the third median.
Dem.—Produce AF to G; draw BG || to EH, and join GC.
Now (xl., Ex. 3) AG is bisected in H; and in the Δ AGC, HD is
, U to GC (xl., Ex. 2); henceBHCG is aO; and .·. (χχχπτ.,Εχ. 1)
BC is bisected by HG, in F. Hence AF is a median of the Δ ABC.
5.	See “Sequel to Euclid,” Book I., Prop, iv., Cor.
6.	Let a, b be the two sides, and c the median of the third side.
It is required to construct a Δ having two sides respectively equal
to a and b9 and the median of the third side equal to c.
Sol.—Construct the Δ ABC, having AB = a, AC = b, and BC
= 2c. Bisect BC in D. Join AD, and produce it until DE = AD.
Join EC. ACE is the required triangle.
Dem.—The Δ8 ADB, CDE are equal (iv.) in every respect;BOOK I.]
EXERCISES ON EUCLID.
37
.·. AB = CE; but AB — a; .·. CE = a, and AC = ft, and BC = 2c;
·. CD = e.
7.	(1) See (xx., Ex. 9).
(2) Let λ, by c be the sides of the Δ, and a, 0, 7 the medians.
Bern.—|i3+f7>a (Ex. 5). Inlike manner %y +f«>ft; and
f α + f β > c. Adding, we hive f (a + β + 7) > (a + ft + e) ; and
therefore (α -f β + y) >f (a+ ft + c).
8.	Let a be the side, and by c, the medians. It is required to
•construct a Δ, having a side equal to a, and the medians of the
remaining sides equal to ft, e.
Sol.—Construct a Δ ABC (xxh.), having BC (the base) = a,
AB = f b9 and AC = §e. Bisect BC in D. Join DA, and produce
to E, so that AE = 2 AD. BEC is the required Δ.
Dem.—Produce BA, CA to meet CE, BE in F, G. Now
ED is a median of the Δ EBC (const.), (4) BF, CG are
medians; hence (5) BA = f BF ; but BA = %b; BF c= b.
Similarly, CG = c.
9.	Let a, ft, e be the medians of a Δ. It is required to con-
struct it.
Sol.—Construct a Δ ABC, having AB = j<*, BC =fft, and CA
= § e. Bisect BC in D. Join AD, and produce it to E, so that
DE=AD. Produce CB to F, and make BF=BC. Join AF,
EF. AFE is the Δ required.
Dem.—Join EB, and produce it to meet AF in H. Produce
AB to meet EF in G. Join CE. Now since AD = DE, and BD
«CD, ABEC is a parallelogram; BH is [j to AC. Hence
(xl., Ex. 3) AF is bisected in H. Similarly, FE is bisected in G,
and (const.) AE is bisected in D; .·. (Def.) AG, DF, EH are the
2a
.medians; hence (Ex. 5) AB = 2BG; but AB = ~; .·. AG s? a.
3
In like manner it can be shown that FD = ft, and EH = e*
10.	Let ABC be the Δ. Let fall a JL AD on BC. Bisect the
Z BAC by AE, meeting BC in E. It is required to prove the
>1 DAE = i (ACB - ABC).
Dem.—From AB cutoff AF = AC. Join CF, cutting AD, AE
iin G,H. Join EF. Now (v.) the L AFC = ACF, and (iv.) the base
,EC = EF; .·. the L EFC = ECF ; hence the L AFE = ACE ;
that AFE = FBE + FEB (xxxii.) ; .*. ACB = ABC + FEB;
Ihmce FEB = ACB - ABC ; but ECF = i FEB ; ECF
= i (ACB - ABC). Again, the L AHG is right (iv., Ex. 1),
.and GDC is right, and the L AGH = CGD (xv.); .-.the L GAH
= GCD. Hence GAH = J (ACB - ABC).EXERCISES ON EUCLID.
[book I*.
11.	Let AM, BN be the two || lines, and P the given point..
It is required to find in AM, BN two points equidistant from P*
and whose line of connexion shall be || to a given line MN.
Sol.—From P let fall a X PQ, on MN. Bisect the part Cl>
between AM, BN in E. Through E draw AB || to MN. A, Ifc
are the required points.
Dem -Join AP, BP. Now the L PEB = PQN (xxix.) j but
PQN is a right L, .·. PEB is right; and since CD is bisected in
E, (xxrx., Ex. 4) AB is bisected in E. Now AE = BE, and.
EP common, and the L AEP = BEP; .·. (τν.) AP =5 BP.
12.	Let a be the side, and b, 0 the two diagonals.
Sol.—Construct the Δ AEB, having AB = a, AE =iji, and
BE — Produce AE, BE to C, D, so that CE == AE* and DE
= BE. Join CD, AD, BC. ABCD is the required parallelogram.
Dem.—The side AB = «, and AC, BD — bte.
13.	Let ABC be a Δ, having the side AB greater than AC..
It is required to prove that BE, the median of AC, is greater
than CF, the median of AB.
Dem.—Let BE, CF intersect in G. Join AG, and produce it
to meet BC in D. AD is the median of BC. Now because BI>
= CD, AD common, and the base AB greater than AC, (xxv.};
the L ADB is greater than ADC. Again, BD = CD, GD com.·
mon, and the L BDG greater than CDG; (xxiv.) BG is
greater than CG; but BG = f BE, and CG = f CF (6). Hence*
BE is greater than CF.
14.	Let AB, CD be two || lines, and E a given point.. It ie
required to find in AB, CD two points that shall subtend a right
angle at E, and be equally distant from it.
Sol.—From E let fall a ± EF on AB. Draw EG f| to AB, and
make it equal to EF. From G draw GH JL to CD,. In AB,taker
FJ=GH. H, J are the required points.BOOK I.]
EXERCISES ON EUCLID.
39
Bern.—Join EH, EJ. Because EF=EG, and FJ =* GH, and
the L EFJ = EGH,	(it.) EJ = EH, and the L FEJ = GBH.
To each add the L FEH, and we have the L JEH = FEG; hut
FEG is a right L. Hence JEH is right.
15.	Let ABC he an isosceles Δ, and D a point in the base BC.
From D let fall X8 DE, DFon AB, AC. From B let fall a X BG
on AC. It is required to prove that BG = DE -f* DF.
Dem.—From D draw DH (j to AC, meeting BG in H. Now
(xxix.) the L HDB = ACD; hut ACD = ABD (hyp.); .·. HDB
= EBB, and the L BHD = BED, each being light; .·. (xxvi.)
BH = DE; hut HG = DF (xxxiv.). Hence BG = DE + DF.
16.	See Book IV., Prop, y., Ex. 1, second proof.
17.	Let ABC he the Δ. Bisect the L BAC by AD, meeting
BC in D. From D draw DE, DF || to AB, AC. AEDF is an
inscribed lozenge.
Dem.—The L EAD = ADF (xxix.); butEAD=FAD (const.);
ADF = FAD, and AF = DF. Similarly, AE = DE; hut
(xxxiv.) AF = DE, and AE = DF. Hence the four sides AF,
DF, AE, DE are equal; .·. AEDF is a lozenge.
18.	See “ Sequel to Euclid,” Book I., Prop. xiv.
19.	(1) Let AB, AC he two fixed lines, and P the point. Let
fall Xs PD, PE on AB, AC; then, being given the sum of PD
and PE, it is required to find the locus of P.
Dem.—Produce EP to F, and make PF = PD. Through F
draw GF || to AC, meeting AB in G. Join GP, and produce it
both ways; GP is the required locus. Because PF = PD, to
each add PE, and we have EF = PD + PE; .·. EF is given;
and since GF is at a given distance from AC, GF is given in
position. Again, since each of the L8 PFG, PDG is right, PF2
+ FG2 = PD2 + DG2; hut PF2 = PD2 (const.); .·. GF2 = GD2;
.·. GF = GD. Now GF = GD, GP common, and the base PF
= PD ; .·. (vm.) the L PGF = PGD. Then, since AB, GF are40
EXEBCISES ON EUCLID.	[book i.
two fixed lines, and GP bisects the L between them, .·. GP is
given in position, and is the locus of P.	,
(2) May be done in like manner.
20.	Let ABC be an equilateral Δ, and P any point within it.
From P let fall X8 PD, PE, PF on AB, BC, CA, and from A let fall
a 1AK on BC. It is required to prove that PD+PE-f PF=AK.
Dem.—Through P draw GH || to BC, meeting AB, AC, AK in
G, H, L; and from G let fall a X GJ on AC. Now the L AGH
=r ABC (xxix.); .*. AGH is an £ of an equilateral Δ. Similarly,
AHG is an L of an equilateral Δ. Hence AGH is an equilateral
Δ; .·. AL = GJ ; but GJ = PD + PF (Ex. 15); .·. AL = PD
+ PF, and PE = LK. Hence AK = PD -f PE + PF.
21.	See “ Sequel to Euclid," Book I., Prop. xi.
22.	See “ Sequel to Euclid,” Book I., Prop, xi., Cor. 1.
23.	Let ABC be a Δ, and L a given length. It is required to
find a point F in BC, such that if FK, FG be drawn || to AB, AC,
the sum of AG, AK shall be equal to L.
Sol.—From B draw BD || to AC, and make it = L. From DBOOK I.]
EXBBCISES ON EUCLID.
41
draw DE j| to AB, and produce AC to meet it in E. Bisect
the L AED by EF, meeting BC in F. F is the point required.
Dem.—Through FdrawGH || to BD, and FK || to AB. Now
the L HEF = EEF (const.), and (xxrx.) the L KEF * EFH ;
.·. EFH = HEF, and .·. HE = HF; but HE = FK, .·. FK
— FH. To each add FG, and we have FK + FG = GH; that is,
AG + AK « GH; but GH = BD = L. Hence AG + AK = L.
24. (1) Let BAC, EDF be two L\ whose legs AB, DE, AC,
DF are respectively ||. Bisect BAC, EDF by AG, DH. It is
required to prove that AG, DH are ||.
Dem.—Join AD, and produce it to J. Now (xxix.) the L JDE
= JAB, and JDF = JAC ; .·. FDE = CAB; hence FDH = CAG.
And it has been shown that JDF=JAC; J DH=JAG. Hence
(xxvui.) DH is parallel to AG.
(2) Let BAC, EDF be the L *. Bisect BAC, EDF by AG, DH.
Produce GA, HD to meet in L. It is required to prove that HL
is ± to GL.
Dem.—Produce FD to J, and bisect the L JDE by DK. Now
the L FDH= EDH, and JDK =* EDK; hence HDK = half sum
of JDE and EDF; but JDE and EDF = two right L8;	HDK
is a right L, and HDK = HLG; .*. HLG is right. And hence
HL is A to GL.
25. Let ABC be the Δ of which A is the vertex; produce BA,
CA to D, E. Bisect the L8 CAD, BAE, by the line FG. From42
EXERCISES OX EUCLID.
[book I.
B, C let fall X* BO, CF, on OF. Bisect the / BAC by AH.
Join BF, CO. It is required to prove that BF, CO meet on
AH.
Dem.—Produce CF to meet AD in D. Now the L CAP
« DAF, and CFA = DFA, and AF is common; (xxvr.) CF
= DF; and because the L DFA = HAF, each being right, AH
is || to CD. Now, since F is the middle point of the base CD*
of the Δ CBD, and BF joined, and AH || to CD, (xxvm.,
Ex. 7), BF bisects AH. In like manner CO bisects AH. Hence*
BF, CO meet on AH.
26. Dem.—From the vertices A, B, C, of the Δ ABC, let fall
Xs AD, BE, CF on the opposite sides; let them intersect in G.
Join DE, EF, FD. It is required to prove that the X» AD, BE,
CF bisect the L · EDF, DEF, and EFD.
Now the L CDE = COE (xxxii., Ex. 5), and BDF = BOF;
but (xv.) CGE = BOF; CDE = BDF; and CDA = BDA^
since each is right; EDA = FDA; hence the L EDF is
bisected by AD. In like manner the L8 DEF, EFD are bisected
by BE and CF.
27· Let ABC be a given Δ, and D a given point within it.
It is required to inscribe, in ABC, a O whose diagonals shall
intersect in D.
Sol.—Through D draw DE || to AC, and from BE cut off FK
= EC. Join FD, and produce it to meet AC in G. Draw DH
II to AB; and from HC cut off HE = BH. Join ED, and pro-
duce it to meet AB in L. Join OL, FL, OE. OLFE is the
required parallelogram.
Dem.—FO is bisected in D (xl., Ex. 3). Similarly, EL is
bisected in D. Hence (xxxiy., Cor. 5) OLFE is a parallelo-
gram.EXEBCISES ON EUCLID.
43
28.	Let aj, «2, «3, *4 be the sides of the quadrilateral; and A, B
the middle points of two opposite sides. It is required to con-
struct it.
80L—Join AB, and on it describe the A ACB, having BC =*
ind CA = § «3. Complete the CUABCD. Join DC; and describe
the A CDE, having ΒΕ»£β2, and CE= \ Complete the
--------
---------H
-----------— s.
CU DECF. Through A, E, B, F draw GH, GK, JK, JH {] re-
spectively to DE, BC, CE, CA. GHJK is the required quadri-
lateral.
Dem.—HF = AC (xxxiv.), and JF = BD; but AC + BI>
* 2 AC; hence H J = «3. In like manner GH = «2, GK = «1, and
JK = 81.
29.	See “ Sequel to Euclid,” Book I., Prop. vm.
30.	Let ABC be the given rectilineal figure, and 0 the given
point. From 0 let fall X· on BC, CA, AB; and let them be
denoted by p, pi, pz; then, if p+pi + Pt be given, it is required
to prove that the locus of 0 is a right line.
Dem.—In BC take a part EF, equal to any given line. Join
OE, OF. In AC, AB take GH, JK, each equal to EF. Join OG,
OH, OJ, OK. Now let EF be denoted by £, and we have
bp = 2Δ OEF (II. 1. Cor. 1), and, similarly, for the Δ8 OGH,
OJK. Therefore b (p -f pi 4- p%) is equal to twice the sum of the
areas of those triangles; but the bases, and sum of the areas, are
given. Hence (Ex. 29) the locus of 0 is a right line.
31.	Dem.—Through C and B' draw CD, B'D || to BB' and
BC. Join DC', cutting BC in E. Now (xxxiv.) BB' = CD?
but BB' « CC' (hyp.); .·. CD = CC', and CE is common; and
the L ACB = DCB, because each is equal to ABC; hence (nr.)
the L CEC' = CED ; .·. each is a right L; .·. (xxix.) B'DE isEXERCISES ON EUCLID.
44
[book I.
right; hence B'C'D is acute; and .·. (xnc.) B'C' is greater than
B'D; that is, greater than BC.
32.	(1) Bern.—From B let fall a ± BC on L; and produce it
to meet AP in Q. In L take any other point S. Join AS, BS,
QS. Now, because BCP = QCP, and the L BPC= QPC, and
CP common, (xxvi.) BP = QP. Similarly, BS *= QS. Hence
AS — SQ = AS — SB; but AS — SQ is less than AQ; .·. AS
— SB is less than AQ; that is, less than AP — BP.
(2) See “ Sequel to Euclid,” Book I., Prop. xxt.
33.	Let ABCD be a quadrilateral. It is required to bisect it
by a line drawn from A, one of its angular points.
Dem.—Join AC. Produce DC to E. Through B draw BE || to
AC. Join AE. Bisect DE in F. Join AF. AF bisects ABCD.
Now the Δ AEC = ABC (xxxyii.) To each add the A ACD,
and we have the Δ AED = the quadrilateral ABCD ; but AED
= 2ADF (xxxviii.) ; .*. ABCD = 2ADF.
34.	Dem.—Bisect ED in F. Join AF. Now (χπ., Ex. 2),
vthe lines EF, AF, DF are equal; hence the L FAD = FDA;
D
JL
B
but (xxxn.) the L AFE = FAD + FDA; .·. AFE= 2FDA, and
.·. (xxix.) = 2DBG; but AF = AB, because each is equal to
\ ED; the L ABF = AFB; but AFB = 2DBC. Hence
ABF = 2DBC.
35.	Dem.—The three L8 ABC, BCA, CAB are equal to two
right L8; .*. ABO, BAO, BCO are equal to a right L; but BOD
= ABO + BAO ; .*. BOD and BCO equal a right L; and EOC
+ BCO equal a right L; hence BOD + BCO = EOC + B'CO;
.·. the I. BOD = EOC.
36.	The angles of each external triangle are respectively equal
to J (A + B), J (B + C), J(A+ B). See (xxxii., Ex. 14). Hence
vthe three external triangles are equiangular.
37.	(1) Dem.—Let ABCD be the quadrilateral* Bisect the L *
iBCD, CDA by CE, DE. It is required to prove that the L CEP
=« J (DAB + ABC).book i.]	EXERCISES ON EUCLID.	45
Now the L 8 DAB, ABC, BCD, CDA are together equal to
four right L8, and the L· CED, EDO, DCE are equal to two
right L *j hence (CED + EDC 4- DCE) = ^ (DAB 4- ABC 4· BCD
4- CDA); but EDC « £ ADC, and DCE = J DCB. Hence CED
= i (DAB 4- ABC).
(2) Bisect the L8 ABD, ACD by BE, CE. Produce BE, CE
to meet AC, BD in F, G. It is required to prove that the L CEF
= J (BAC - BDC).
Dem.—Join AE. Now the L8 of the figure ABEC are equal to -
four right L8; and the L8 of the figure BECD equal to four right
Z8» hence the Z.8 (BAC 4· ABE 4· BEC 4· ACE) = (BEG 4* GEF
4- FEC 4- ECD 4- CDB 4- DBE); hut ABE = DBE, and ACE
= ECD, and BEC *= GEF. Reject these, and we have BAC
= CDB 4- GEB 4- CEF = CDB 4- 2 CEF. Hence the L BAC
exceeds CDB by 2 CEF; that is, CEF = \ (BAC - CDB).
38.	Dem.—It has been proved (xlvii., Ex. 7) that EF2 = AC2
4- 4 BC2. Similarly, EG2 = BC2 4- 4 AC2. Adding, we get EF2
4- KG2 = 5 (AC2 4- BC2) = 5 AB2.
39.	Let A, B, C, D, E he the middle points of the sides of a
convex polygon of an odd number of sides. It is required to
construct it.
Sol.—Join CD, DE; and through C, E draw CF, EF || to DE,
CD; and (xxxiv., Ex. 6) construct the Δ GHJ, having A, B, F
for the middle points of its sides. Join AF, BF, JC, and pro-
duce JC to E, so that CE = CJ. Join ED, HE, and produce
them to meet in L. GHLEJ is the required polygon.
Dem.—Join HE. Now in the Δ HJE, HJ, JE are bisected
in F, C; hence (xl., Exercises 2 and 5) FC is || to HE, and46
EXERCISES ON EUCLID.
[book t.
«qua! to half of it; but EC = ED; ED is || to HK, and
equal to half HK. And hence (xl., Ex. 3) HL, LK are bisected
in E, D.
40. Let ABCD be a quadrilateral. It is required to trisect it
by lines drawn from C, one of its angular points.
Sol.—Join BC. Produce DB to E, and draw AE || to BC.
Join CE. Trisect ED in F, G (xxxiv., Ex. 3). Join CF, CG.
CF, CG trisect the quadrilateral.
Dem.—The Δ CEB = CAB (xxxvii.). To each add CBD,
and we have the Λ CED = the quadrilateral CABD; but the
Δ CGD * J CED; .·. CGD = i CABD. In like manner CFG
= iCABD.
41. Let ABC be a Δ whose base BC is given in magnitude and
position; and the sum of its sides BA, AC also given. Produce
BA to D, and make AD * AC. Bisect the L CAD by AE.
Erect CE X to AC. Join BE, DE; and from E let fall a X EF
on BC produced It is required to prove that the locus of E is
the perpendicular EF·
Dem.—Because AC = AD, and AE common, and the L CAEBOOK I.]
EXERCISES ON EUCLID.
47
-e DAE, .·. (iv.) CE = DE, and the L ACE = ADE; but ACE
is a right L (const.); ADE is right; hence (xlyii.) BE2
— ED2 = BD2; but BD is given, since it is equal to BA + AC ;
•and ED =EC ; .*. BE2 — EC2 is given, and the base BC is given.
Hence (xlvii., Ex. 5) the locus of E is EF, the i. from E
on BC.
42. (1) See xxxii., Ex. 8.
(2) Let ABCD he a parallelogram. It is required to prove
that EFGH is a rectangle.
Bern.—The L* ABC, BAD are together equal to two right L »
(xxix.); .·. the U EBA, EAB together make a right L ; hence
the L AEB is right. Similarly, the L8 at F, G, H are right.
Hence EFGH is a rectangle.
(3) Let ABCD be a rectangle. It is required to prove that
EFGH is a square.
Dem.—Because the L BAD = CDA; the L BAE = CDG.
In like manner the L ABE = DCG, and the side AB = CD;
.·. (xxvi.) AE = DG; but AH = DH, since the L ADH = DAH;
HE = HG. In like manner all the sides are equal, and
the L* are right L8. Hence EFGH is a square.
43.	Dem.—Join AE. Now (xl., Ex. 6) EF = J AB = BD;
and FG = BD ; .·. EF = FG, and AF = CF (hyp.) ,* .*. CF and
FG = AF, FE ; and the L CFG = AFE (xv.); hence (iv.) CG
= AE; but AE is a median of the Δ ABC; also CD, a side of
the Δ CDG, is one of the medians of ABC; and BF, the remaining
median, is equal to DG (xxxiv.)· Hence the sides of the Δ CDG
are equal to the medians of ABC.
44.	Let ABCD be the billiard table, E the point from which
the ball starts, and F the point through which it will pass.
Sol.—From E let fall a JL EG on AB ; produce EG to H, so
that GH = EG. From H let fall a _L HJ on CB produced; and
produce HJ to K, so that JK = HJ. From F let fall a X FL on
AD, and produce to M, so that LM = LF; and from M let fall aEXERCISES ON EUCLID.
[book I..
_L IfN on CD produced, and produce to P, so that NP=MN. Join
XP, intersecting BC in Q and CD in B. Join HQ, MB, inter-
be the path of the ball.
Dexn.—Because EG=HG, GS common, and the L EGS=HGSr
.·. the L ESG = HSG; but HSG = BSQ (xv.); .·. ESG=BSQ;
hence the ball will he reflected in the direction SQ. In like man-
ner it can be shown that the L HQJ = BQC, and therefore the
ball will be reflected from Q in the direction QB. Similarly, it
will be reflected from B to BT, and from T to TF.
45. Let ABC be the Δ, AD, BE the bisectors of the L * A, B.
It is required to prove, if AD = BE, that the L CAB = ABC.
Bern.—If the angle CAB be not equal to ABC, let CAB
be the greater; then, since the L CAB is greater than ABC, its
half, the L DAC, is greater than EBC, the half of ABC; then
make DAF equal to EBC. Now, since the L DAB is-greater
than ABE, the whole L FAB is greater than FBA; .-.the side
FB is greater than FA. Cut off BG = FA, and draw GH parallel
to FA; then the Δ* GBH, FAD have evidently two angles in
one respectively equal to two angles in the other, and the side
BG = AF. Hence BH is equal to AD ; hut BE is = AD (hyp.).
Hence BH = BE, which is absurd» Hence the angle CAB is not
unequal to ABC; that is, it is equal to it, and (vi.) theΔ ABC
is isosceles.book i.]	EXERCISES OK EUCLID.	49
46. Let ABC be a Δ, whose base and difference of sides are
given. Bisect the L BAG by AD. Erect CD 1 to AC* The
locus of D is a right line.
Dem.—Let fall a A DE on AB. Join BD. Now (I. xxvi.)the
A*ACD, AED are equal in every respect; DC = DE, and
AC=AE; .·. AB-AC=*BE; but AB - AC is given; .*.BEia
given. Again, BD3 - DE3 = BE3; that is, BD3 - CD3 = BE3;
hence BD3 - CD3 is given, and the base BC is given. Now we
axe given the base, and the difference of the squares of the sides
of the Δ BCD. Hence (xlvii., Ex. 5) the locus of the vertex
D is a right line perpendicular to the base.
47. Let EFGH be a square inscribed in the Δ ABC. It is
required to prove that (BC + AD) a = 2 Δ ABC, where a denotes
the side of the square.
Dem.—Let fall a A AD on BC. Join DF, DG. Now BD.EF
= 2 Δ BFD (II. i., Cor. 1); that is, BD. a = 2 Δ BFD. Simi-
larly, DC. s = 2 Δ DGC; BC. a = 2 Δ BFD + 2 Δ DGC.
Again, AD.FK a 2 Δ AFD, and AD.GK = 2Δ AGD; AD .a
= 2 AFDG. Adding, we get (BC + AD) a = 2 Δ ABC.EXERCISES ON EUCLID.
[book I.
60
48.	Dem.—Let fall a A CE on AB. Now (xlvii., Ex. 20)
BC2 = AB.BE + AC.CD ; but (xxvi.) the Δ· BEC, BDC are
equal; since the Δ ABC is isosceles; .*. BE — DC, and AB = AC.
Hence BC2 = 2 AC. CD.
49.	Let ABC be a right-angled Δ, and let equilateral Δ® be
described on its three sides. It is required to prove that the
Δ ABD is equal to the sum of the Δ· ACF, BCE.
Dem.—Bisect AC in G. Join FO, BG, FB, CD. Now the
L CAF = BAD ; to each add CAB, and we have the / FAB
= CAD, and AF = AC, and AB = AD ; .\(iv.) the Δ* AFB,
ACD are equal. Again, because each of the Z* FGC,ACB is
right, BC, FG are parallel; .·. (xxxvn.) the Δ FGC = FGB. To
each add the Δ FGA, and we have AFC = to the quadrilateral
AFBG. Again, to each add the Δ AGB, which is £ ACB, and
we have AFC + £ ACB = AFB. Hence ACD = AFC + £ ACB.
Similarly BCD = BEC 4- £ ACB. Add, and we have ACBD
= AFC -f ACB + BEC. Reject the right angled Δ ACB, which
is common, and the Δ ABD = AFC + BEC.
60. (1) Let AB be the base, X the difference of the base Z_*,
and S the sum of the sides. It is required to construct the
triangle.
Sol.—Draw BD, making the L ABD = £ X, and draw BC
X to BD. With A as centre, and a radius equal to S, describe a
0, cutting BC in C. Join AC, cutting BD in E. Bisect CE
in F. Join BF. AFB is the required triangle.
Dem.—The lines BF, CF, EF are equal (in., Ex. 2); .·. FEβοοχ I.]
EXEBCISES ON EUCLID.
61
« PB; .·. the LFBE = FEB; but FEB = FAB + ABE (xxxit.) ;
FBE = FAB + ABE; beuce the l_ FBA — FAB + 2 ABE)
and hence theV ABE is half the difference of the base L*; but
ABE = J X. Hence the difference of the base Z.· = X; and since
FB = FC, AF + FB = AC = S, .·. the sum of the sides = S.
(2) Let AB be the base, X the difference of the base L%, and D
the difference of the sides.
Sol.—Draw BE, making the L ABE = J X. With A as centre,
and a radius equal to D, describe a Θ, cutting BE in E. Join AE,
A	JB
and produce it. Draw BC, making the L CBE = CEB, and meet-
ing AE produced in C. ACB is the required triangle.
Dexn.—CB = CE (vi.)) AE — AC — CB) but AE = D j
.·. AC — CB = D, and, as before, the difference of the base angles
= X.
51. Sol.—Let AB be the base, and M the median that bisects
the base. To AB apply a CD ABCD, whose area is equal to twice
the given area (xlv.). Bisect AB in E. With E as centre,
and a radius equal to', M, describe a Θ, cutting CD in F. Join
AF, BF. AFB is the required Δ.
52. Deni.—Join AO, CG, FO. The Δ CED = CGD + CEG,
and the Δ EBC = BGC - CEG. Subtracting, we get CED - EBC
έ 252
EXEBCISES ON EUCLID.
[book I.
= 2 CEG. Similarly AED — AEB = 2 AEG. Subtracting, we have·
AEB + CED - (AED + EBC) = 2 (CEG - AEG). Again, CEG
* CFG + EFG, and AEG = AFG - EFG; .·. CEG - AEG
= 2 EFG. And hence 4 EFG — AEB 4· CED — (AED + EBC)..
53. (1) Let ACB be the Δ. Describe squares AH, AF, CE
on the sides AC, AB, BC respectively. Bisect AC in J. Joint
BJ, EF. It is required to prove that EF = 2 BJ.
Dem.—Produce BJ to M, so that JM = JB, and join MC.
Now (iv.) the Δβ MJC, AJB are equal in every respect;
MC = AB = BF, and CB = BE; hence MC, CB equal:
BF, BE. And because AC and BM bisect each other in J, MC
and AB are parallel; .*. the Z" MCB and ABC are together equal
to two right Z% and the Zs EBF, ABC are equal to two right Z*;
since ABF and CBE are right; .*. the L MCB = EBF; hence
(iv.) MB = EF; hut MB = 2 BJ; EF = 2 BJ.
(2) Produce MB to meet EF in N. MN is JL to EF.
Dem.—From the equal triangles CMB, BFE we have the
L CMB = BFE, but CMB = ABM; .·. BFE = ABM. To each
add NBF; and we have BFN + NBF = ABM -f NBF; hut
since ABF is right, ABM + NBF equal a right Z; .*· BFN
+ NBF equal a right Z; and hence the L BNF is right.BOOK Π.]
'EXERCISES ON EUCLID.
BOOK II.
PROPOSITION IV.
. 1. Dem.—AB2 = AB. AC + AB . BC (n.);
but	AB. AC = AC3 -I- AC . CB	(iii.) ;
and	AB.BC = BC2 + AC.CB	(iu.) ;
Therefore AB. AC + AB. BC = AB2 + BC2 + 2AC . CB;
that is, AB2 = AC2 + BC2 + 2 AC . CB.
2.	Let C he the vertical L of the right-angled A ABC. From
*C let fall a _L CD on AB. It is required to prove that DC2
= AD. DB.
Dem.—AB2 = AC2 + CB2 (I. xlvii.) ; hut AC2=AD2 + DC2;
and CB2 = BD2 + DC2; .·. AB2 = AD2 + BD2 + 2 DC2. Again,
AB’ = AD2 + DB* + 2 AD . DB (iv.). Hence DC2 = AD. DB.
3.	Let ABC he the right-angled Δ. In the base AB cut off
AD = AC, and BE = BC. It is required to prove that ED2
= 2 AE.DB.
Bern.—AB2=AC2 + CB2 (I. xlvii.) = AD2 + BE2; but AD2
*	AE2 -l- ED2 + 2 AE. ED (iv.); and BE2 = BD2 + DE2
+ 2 BD.DE; .·. AB2 = AE2 + ED2 + 2 AE . ED + BD2 + DE2
+ 2 BD. DE; also AB2 = AE2 + ED2 + DB2+2 AE. ED + 2 ED.DB
*	2 AE . DB (iv., Cor. 3). Hence ED2 = 2 AE. DB.
4.	Let ABC he the right-angled Δ, CD the X from the right
angle on the base. It is required to prove that (AB + CD)2 ex-
ceeds (AC + CB)2 by CD2.
L em.—AC . CB is equal to twice the Δ ACB, and AB . CD
is equal to twice the Δ ACB ; .·. AC . CB = AB . CD.
No*	(AB + CD)2 = AB2 + CD2 + 2 AB . CD;
rand	(AC + CB)2 = AC2 + CB2 + 2 AC . CB.
Subtracting, we have (ΑΒλ+ CD)2 — (AC + CB)2 = AB2 — BC254	EXERCISES ON EUCLID.	[book it.
- CA2 + DC2; but AB2 - BC2 = AC2 ; . \ (AB + CD)2 - (AC + CB)3
=AC2 - AC2 + DC2 = DC2.
5. Let the sides of the Δ he denoted by a, l, e, e being t
hypotenuse. It is required to prove that (a + b + c) — 2 (e + a)(c+ l·)-
Bern.—{a + b + c)2 = a2 + IP + c* + 2ab + 2ac + 2bc\ but(a2 + δ2
= c* (I. xlvii.) ; .*. (a2 + bz + <*) = 2c*. Hence (a + b +
= 2(<S* + ae +	+ ab)- 2 (c + a) (e + δ).
PROPOSITION Y.
1.	Let AB be the given straight line. Bisect it in C. It is
required to prove that AC . CB is a maximum.
Dem.—Take any other point D in AB; then AD . DB + CD2
= CB2 (v.); but CB2 = AC . CB ; .·. AC . CB = AD.DB + CD2 ;
that is, AC. CB is greater than AD . DB by CD2. Hence, when,
a line is bisected, the rectangle contained by the parts is a maxi-
mum.
2.	Let AB be the given straight line, and L the line whoso
square is given. It is required to divide AB, so that the rect-
angle contained by its segments will be equal to L3.
Sol.—Bisect AB in C ; with C as centre, and CB as radius,,
describe a semicircle. Draw BD X to AB, and = to L. Through.·
D draw DE || to AB, cutting the semicircle in E; let fall a X EF
on AB. The rectangle AF. FB = L2.
Dem.—Join CE. Now AF.FB + CF2 = CB2 (v.) = CE*
= CF* + FE2 (I, xlvii.). Take away CF2, which is common,
and AF. FB * FE2 = BD2 = L*.
3.	Let ABC be the Δ. From C let fall a X CD on AB. It is
required to prove that (AC + BC) (AC - BC) = AB (AD - DB).
Dem.—AC* = AD2 + DC* (I. xlvii.) ; and BC2 = BD2 + DC2.
Subtracting, we get AC2 - BC2 = AD2 - DB2 ; that is (AOBOOK II.]
EXERCISES ON EUCLID.
55
+ BC) (AC - BC) = (AD + DB) (AD - DB) = AB (AD
- DB).
4.	Dem.—(AC + BC) (AC - BC) = AB (AD - DB) (Ex. 3);
but (AC + BC) ie greater than AB (I. xx.); .·. (AC - BC) is less
than (AD — DB).
5.	Let ABC be an isosceles Δ. From C let fall a X CE on AB.
In AB take any other point D. Join CD. It is required to prove
that CB2 - CD2 = AD . DB.
Dem.—AD. DB + ED2 = EB2, and EC2 = ED2. Add together,
and we get AD.DB + CD2 = CB2 ; .·. AD. DB = CB2 - CD2.
6.	Let ABC be the Δ. It is required to prove that AC2
= (AB + BC) (AB - BC).
Dem.—AC2 + BC2 = AB2; .·. AC2 = AB2 — BC2 = (AB
+ BC) (AB - BC).
PROPOSITION YI.
1.	Let AD be the straight line which is bisected in C, and
divided unequally in B. It is required to prove Prop. vi. by
Prop, v., by producing the line DA in the opposite direction.
Dem.—Produce DA to 0, and make OA = BD.
Now OB. BD + CB2 = CD2 (v.); but since OA = BD, OB
= AD. Therefore AD. DB -f CB2 = CD2.
2.	Let AB be the given line. It is required to divide it exter-
nally in E, so that AE. EB = L2, L being a given line.
Sol.—Bisect AB in C (vi.). Erect BD X to AB, and make it
equal to L. Join CD. With C as centre, and CD as radius,
describe a circle, meeting AB in E. E is the point required.
Dem.—Now AE.EB + CB2 = CE2 = CD2 = CB2 + BDV:
Reject CB2, which is common, and AE. EB = BD2 = L2.
3.	See Ex. 2.
4.	Let AD, DB be two lines. Bisect AB in C.
Dem.—Because AB is the sum, CB is half sum; and AD = AC
+ CD, and DB = CB — CD; .·. AD — DB = 2 CD ; hence CD is
half difference. Now AD . DB + CD* = CB2 (v.); .·. AD. DB
= CB* - CD2.
5.	Dem.—Let AB be the sum, and D2 the difference of their
squares. To AB apply the rectangular □ ABCE = D2. Now,
since the sum multiplied by the difference is equal to the differ-
ence of the squares, and that AB is the sum, therefore AE must be
the difference. Produce BA to F, and make AF = AE· Therefore,$6
EXERCISES ON EUCLID.
[book u.
since the sum together with the difference is equal to twice the
greater; therefore if we bisect BF in G, BG will be the greater,
and AG the less.
If we take ΑΈ equal to the difference, and apply the rect-
angular □ ABCE = D2, we have the second case.
F
E,_________________ C
AG	B
D
6.	See “ Sequel to Euclid,” Book II., Prop, i., Cor.
7.	The rectangle contained by two straight lines, together with
the square described on half their difference, is equal to the square
on half their sum.
PROPOSITION VIII.
1.	Dem.—By the third proof of Prop. vm. (AB + BO)2
= 4 AB. BO + AO*; but AB.BO = BC2 (I. xlyii., Ex. 1), and
AO2 = AC2 - CO2; .·. (AB + BO)2 = 4BC2 + AC2 - CO2;
but 4BC2 + AC2 = EF2 (I. xlvii.. Ex. 7); (AB + BO)2
- EF2 - CO*.
2.	Dem.—GK2 = 4 AC2 + BC2 (I. xlyii., Ex. 7), and EF2
= 4 BC2 + AC2; GK2 - EF2 = 3 AC2 - 3 BC2; but (I. xlyii.,
Ex. 1) AC2 = AB. AO, and BC2 « AB.BO; .*. GK2 - EF2
= 3 (AB. AO - AB. BO) = 3 AB (AO - BO).
3.	Sol.—Let AB be the difference of the lines. Bisect AB
in C ; erect BE A to AB, and make it equal 2 AB = 2 R. Join
CE, and produce CB to D. Cut off CD = CE. AD, DB are the
required lines.
Dem*—AD. DB + CB2 = CD2 (vi.) = CE2 = CB2 + BE*.κοοκ π.]
EXERCISES ON EUCLID.
57
Reject CB2, which is common, and we have AD. DB = BE-
= 4 R*. Hence AD, BD are the required lines; for their diffe-
rence is AB, that is, R, and their rectangle is equal to 4 R2.
PROPOSITION IX.
1.	Let AB be the given line. Bisect it in C. It is required to
prove that AC* + CB2 is a minimum.
Bern.—Take any other point D in AB. Now AD* + DB8
= 2AC* + 2CD* (ix.) = AC2 + CB* + 2CD*; therefore
AC* + CB* is less than AD* + DB* by 2 CD*. Hence, when
a line is bisected, the sum of the squares on its segments is a
minimum.
2.	Let AB be a given line. It is required to divide it inter-
nally, so that the sum of the squares on the parts may be
equal to L2.
Sol.—Draw BC, making the L ABC half a right £. With A
a* centre, and a radius equal to L, describe a Θ, cutting BC in D.
Prom D let fall a 1 DE on AB. E is the point required.
f Dem.—Because the L EBD is half a right L, and the L BED
right, the L BDE is half a right L; .*. EB = ED ; .·. EB2
= ED*; AE2 + ED2, that is, AD2, that is L2 = AE2 + EB2.
If the circle does not meet the line BC, the question is impos-
sible.
3.	Dem.—From AC cut off AE = DB. Now AD* + AE2
AD . AE + ED2 (vii.); that is, AD2 + DB2 = 2 AD. DB
+ 4 CD*.58
EXERCISES ON EUCLID.
[book II.
4.	Let AEB be the Δ. In AB take any point D. Join ED.
It is required to prove that 2 ED2 = AD2 + DB2. From E let
fall a 1 EC on AB. Now AD2 + DB2 = 2 AC2 + 2 CD2 (ix.) ;
but AC = CE. Therefore AD2 + DB2 = 2 EC2 + 2 CD2 = 2 ED2.
5.	See “Sequel to Euclid,*’ Book II., Prop. xn.
PROPOSITION X.
1. (1) Let AB be the sum of the lines, and 2 X2 the sum of the
squares.
Sol.—Bisect AB in C. Erect CD A to AB, and make it equal
to AC or CB. Produce DC to E. Cut off DE = X. With D as
centre and DE as radius, describe a Θ, cutting AB in F. AF
and FB are the required lines.
Dem.—Join DF, DB. Now AF2 + FB2 = 2 AC2 + 2 CF-
(ix.) = 2 DC2 + 2 CF2 = 2 DF2 = 2 DE* = 2 X2.
D
(2) Let AB be the difference, and 2 X2 the sum of the squares.
Sol.—Bisect AB in C, and erect CD A to AB, and make it
equal to AC or CB. Produce DC to E. Cut off DE = X. With
D as centre, and DE as radius, describe a Θ, cutting AB pro-
duced in F. AF and FB are the required lines.
Dem.—Join DB, DF. Now AF2 + FB2 = 2 AC2 + 2 CF2
= 2 DC2 + 2 CF2 = 2 DF2 = 2 DE2 = 2 X2.
2. Let CE be the median which bisects the base AB. It is
required to prove that AC2 -f CB2 = 2 AE2 + 2 CE2.
Dem.—From C let fall a A CD on AB. Now AD2 + DB2
= 2 AE2 + 2 ED2 (ix.), and CD2 + CD2 = 2 CD2. Add, and we
get AC2 + CB2 = 2 AE2 + 2 CE2.BOOK Π.]
EXERCISES ON EUCLID.
59·
3.	Let BC be the given base of a Λ ABC, the sum of the
squares of whose sides AB, AC, is equal to a given square. It
is required to prove that the locus of the vertex A is a circle.
Dem.—Bisect BC in D. Join AD. Now (Ex. 2), BA-
+ AC2 = 2 BD2 + 2DA2; but BA2 + AC2 is given (hyp.);
.·. 2 BD2 + 2 DA2 is given, and 2 BD2 is given, since BD is
half of the given base BC, .*.2 DA2 is given; .·. DA is givenr
and the point D is given. Hence the locus of A is a circle,
having D as centre, and DA as radius.
4.	Bern.—Bisect AD in E. Join BE, CE. Now (Ex. 2)
AB2 + BD2 = 2 AEa + 2 BE2, and AC2 + CD2 = 2 AE2 + 2 GW ;
but AB2 + BD2 = AC2 + CD2 (hyp.); hence 2AE2 + 2 BE2
= 2 AE2 + 2 CE2, and therefore 2 BE2 = 2 CE2; . \ BE = CE.
5.	See “ Sequel to Euclid,” Book II., Prop. in.
PROPOSITION XI.
1.	Let AB be the line. It is required to cut it externally in-
extreme and mean ratio.
Sol.—Erect BC X to and equal to AB. Bisect AB in D. Join
DC. Produce AB to E. Cut off DE = DC. AB is cut in E in
extreme and mean ratio.
Dem.—AE. EB 4- DB2 = DE2 (vi.) = DC2 = DB2 + BC2.
Reject DB2, which is common, and AE. EB = BC2 = AB2.
2.	Let AB be a line divided in extreme and mean ratio at C.
It is required to prove that AC2 — CB2 = AC . CB.
Dem.—AB. BC = AC2 (hyp.); but AB = AC + CB ; .*. (AC
+ CB) CB = AC2; that is, AC. CB -f CB2 = AC2; and therefore
AC . CB = AC2 - CB2.
3.	Let ACB be a right-angled Δ, having AC2 = AB.BC.
From C let fall a X CD on AB. It is required to prove that
AB.BD = AD2.
Dem.—AC2 = AB. BC (hyp.), and AC2 = AB . AD (I. xlvii.,
Ex. 1); .·. AD = BC; .·. AD2 = BC2; but BC2 = AB.BI>
(I. xlvii., Ex. 1). Hence AB. BD = AD2.
4.	(1), Dem.—AB2 + BC2 = 2 AB. BC + AC2 (vn.); but
AB.BC =» AC2 (hyp.) Hence AB2 + BC2 = 3AC2.
(2) Dem.—(AB + BC)2=4 AB. BC + AC2 ( viii.) ; but AB. BC
= AC2 (hyp.) Hence (AB + BC)2 = δ AC2.*0
EXEKCISES ON EUCLID.
[book II.
*5. Dem.—Join FK, AD. Now the square AFGH is double
of the Δ AFK (I. xli.) And the rectangle HBDK is double of
AKD; but AFGH « HBDK (xi.); the Δ AFE = AED; and
hence (I. xxxix.) AK is parallel to FD. In like manner, by join-
ing BF, GD, it can be shown that GB is parallel to FD. Hence
the three lines AE, FD, GB are parallel.
6.	Dem.—Join BF, and produce CH to meet it in L.
Because EB = EF, the L EBF = EFB, and the L * at L are
right (xi., Ex. 7); .*. the L BOL = FCL; hut BOL = EOC;
. ■. EOC — ECO, and . *. EC = ΈΟ ; hut EC — EA j .*. ΈΟ = EA i
.·. the l EOA = EAO, and EOC = ECO. Hence the L AOC
= OAC + OCA, and is therefore (I. xxxii., Cor. 7) a right angle.
7.	Let CH be produced to meet BF at L. It is required to
prove that CH is perpendicular to BF.
Dem.—The Δ· FAB, HAC, are equal (I. it.) in every reject;
.*. the L FBA = HCA, and the L LHB = AHC (I. xv.); the
L HLB = HAC (I. xxxii., Cor. 2); but HAC is a right angle.
Hence HLB is right.
8.	Dem.—In AB take AH = BC — AC. Produce CA to F, so
ithat AF = AH, then evidently CF = CB. Complete the square
AFGH. Produce AC to E, and make CE = AC, and complete
H	G B
	
£	ED
"the square ABDE. Produce GH to meet ED in K. Now we have
the construction as in Prop, xi., and .·. AB. BH = AH2. Hence
AB is divided in “ extreme and mean ratio ” at H.
* See diagram in Euclid [II. xi.] for this and the two follow-
ing Exercises.BOOK II.]
EXERCISES ON EUCLID.
PROPOSITION XII.
1.	Bern.—Produce AC, and let fall a JL BD on AC produced.
Make DE = CD, and join BE. Now the Δ* BCD, BED are
equal in every reepect (I. iv.); .\ the L BCE = BEC. And
since the L ACB is twice an L of an equilateral A, each of the-
A· BCE, BEC is an A of an equilateral A ; hence the A BCE is
equilateral; .·. BC = CE = 2CD.
Again, AB* = AC2 + CB2 + 2 AC.CD; but we have shown that
BC = 2 CD. Hence AB2 = AC2 + CB2 + AC . CB.
2.	Bern.—Join AC; bisect it in 0. Join BO, DO, EO. Now
the lines AO, BO, CO are equal (I. xn., Ex. 2); hence OBC is
an isosceles A ; .*· OE* — OC* = BE.CE (vr., Ex. 6). In like
manner OE* — OD* = AE .DE; but OC = OD. Hence AE. DE
= BE.CE.
3.	Dem.—^Produce AB, DB. Cut off BE =DC, and BF = BC

A
D C*62	EXERCISES ON EUCLID.	[book n.
Join AF, FE. Now the Z FBC = BDC + BCD (I. xxxn.); but
EBC and BDC are right angles; .·. FBE = BCD ; hence (I. iv.)
the Δ· BEF and BDC are equal in every respect; therefore the
Z BEF = BDC ; hence the Z BEF is right. Now AF2 = AB2
+ BF2 + 2 AB.BE (xii.), and AF2 = AB2 + BF2 + 2 BF.BD;
.·. AB . BE = BF. BD; but BE = DC, and BF = BC. Hence
AB. DC = BD . BC.
4.	Dem.-Erect BD 1 to AB, and equal BC. Join AD, CD.
Now AD2 = AC2 + CD2 + 2 AC. CB (xii.), and CD2 = CB2 + BD2
= 2 CB2 = AC2 ; .·. AC2 + CD2 = 2 AC2; .·. AD2 = 2 AC2
+ 2 AC. CB = 2 AC (AC + CB) = 2 AC. AB. Again, AD2 = AB2
+ BD2; but BD = BC; .·. AD2 = AB2 + BC2. Hence AB2
4- BC2 = 2 AC.AB.
5.	Sol.—Bisect AB in 0. From D let fall a J_ DF on AB.
Divide OF in C, so that 2 OC.CF = p2 (the given square). C is
.the point required.
Dem.—Erect CE 1 to AB. Join OE, OD, CD.
Now OD2 = OC2 + CD2 + 2 OC. CF (xii.) = OC2 + CD2 -i- p*;
but OE = OD; .\ OE2 = OC2 + CD2 + ^; that is, OC2 + CE2
= OC2 + CD2 + p*; .·. CE2 - CD2 = pK
6.	Dem.—AD. DB = CD2 - CB2 (vi., Ex. 6); but CD2 * 2 AB2
(hyp.); .·. AD. DB = 2 AB2 - CB2 = 2 AB2 - AB2 = AB2.
PROPOSITION XIII.
1. Dem.—From the vertex A let fall a 1 AD on BC. From
DB cut off DE = DC. Join AE. Now the A ACD = AED
in every respect (I. rv.); .*. AC = AE, and the Z AEC = ACE;
hence AEC is an Z of an equilateral A; the A ACE is
equilateral; .·. AC = CE = 2 CD. Again, AB* = BC2 + CA*
- 2 BC. CD (hi.) ; but we have shown that 2 CD = AC.
Hence AB2 = BC2 + CA2 - BC. AC.BOOK II.]
EXERCISES ON EUCLID.
«3
2.	See “ Sequel to Euclid,” Book II., Prop. rv.
3.	Sol.—Erect BD X to and equal to AB. Join AD. Produce
AB to C. Cut off AC = AD. C is the point required.
Bern.—AD2 = AB2 + BD2 = 2AB2; .·. AC2 = 2AB2. To
*each add BC2, and we have 2 AB2 + BC2 = AC2 + BC2
= 2 AC.BC + AB2 (to.) ; .*. AB2 + BC2 = 2 AC.BC.
PROPOSITION XIV.
1. Sol.—Let a line CD be found (xiv.) whose square is equal
to the given difference of squares. On CD construct a rectangle
CE, equal to the given rectangle. Produce CD to A, so that
CA. AD = DE2 (vi., Ex. 2). Produce ED. From A inflect AB
= DE to the line DB, and join BC. BC and BD are the required
lines.
Dem.—Because AB2 = DE2 = CA. AD, the L ABC is right
(I. klvii., Ex. 1); .\ AB.DC = BD.BC (xn., Ex. 3); hence
the rectangle CE = BD . BC, and CE is equal to the given
rectangle. Also because the L BDC is right, BD2 — BC2 = DC2,
which is equal to the given difference of squares.
2. See Book II., Ex. 6, Miscellaneous.
Miscellaneous Exercises on Book II.
1. Let ABCD be a quadrilateral, AC, BD its diagonals, and
EF, GH lines joining the middle points of BC, AD, AB, CD
It is required to prove that AC2 + BD2 = 2 EF2 + 2 GH2.64
EXERCISES ON EUCLID.
[book n.
Dem.—Join GE, EH, HF, FG. Now GEHF is a parallelo-
gram (I. xl., Ex. 6); .*. 2 GH2 + 2 EF2 = 2 GE2 + 2 EH2
+ 2 HF2 + 2 FG2 (x., Ex. 5) = 4 GE2 + 4 EH2.
Again, GE = } AC (I. xl., Ex. 5), and EH = | BD; .*.4 GE*
+ 4 EH2 = AC2 + BD2. Hence 2 GH2 + 2 EF2 = AC2 + BD2.
2.	Let AD, BE, CF be the medians.
Dem—AB2 + AC2 = 2 BD2 + 2 AD2 (x., Ex. 2);	2 AB2
+ 2 AC2 = BC2 + 4 AD2; but AO = f AD; .·. AO2 = * AD2;
.·. 9 AO2 = 4 AD2 ; hence 2 AB2 + 2 AC2 = BC2 + 9 AO2.
Similarly 2 AC2 + 2 CB2 = AB2 + 9 CO2, and 2 CB2 + 2 AB2
= AC2 + 9 BO2; .*. 3 (AB2 + BC2 + CA2) = 9 (AO2-t BO2 + CO2).
Hence AB2 + BC2 + CA2 = 3 (AO2 + BO2 + CO2).
3.	Sol.—Construct the Δ OCG, having OC = D, OG = 2E,
and CG = F. Bisect OG in B. Join CB, and produce it to A.
Cut off AB = BC. Join AO. OA, OB, OC are the required
line·.
D
Dem.—The Δ» ABO, CBG are equal in every respect (I. nr.)';
.·. AO = CG = F, and OC = D, and OBb= E.
4.	Let ABCD be a quadrilateral; AC, BD its diagonals. Bisect
AB, CD in EF. Join EF. It is required to prove that AD2
-f BC2 + AC2 + BD2 = AB2 + DC2 + 4 EF2.
Dem.—Join CE, DE. Now AD2 + BD* = 2 AW + 2 ED2
(x., Ex. 2), and AC2 + BC2 = 2 BE2 + 2 CE2; .·. AD2 + BD2
+ AC2 + BC2 = 2 AE2 + 2 BE2 + 2 CE2 !+ 2 DE2; but 2 AE2
+ 2 BE2 = 4 AE2 = AB2; and 2 CE2 + 2 DE2 = 4 DF* + 4 EF2
= DC2 + 4 EF2. Therefore AD2 + BD2 + BC2 + AC2 = AB2
+ DC2 + 4 EF2.
5.	Let a, 4, c be the sides of the triangle. On a, b, e describe
squares. Join the adjacent comers, and let the joining lines he'65
book II.] EXERCISES ON EUCLID.
denoted by a, 0, 7. It is required to prove that a2 + jB* + y2
*3 (a2 + 5»+e2).
Dem.—Complete the construction, as in I. xlvii., Ex. 6.
Now we have (x., Ex. 2) aa + a2 = 2 b2 + 2 c2; β2 + b2 m 2 c2
+ 2 a2; and y2 + c2 = 2 a2 + 2 IP. Add together, and we get
a2 + /32 + 72 + (a2 + 52 + c2) = 4 (a2 + 52 + c2); and .*.	+ £2
+ T2 = 3 (α2 + ό2 + c2).
6. Let AB be a given Hne. It is required to divide it into two
parts at C, so that the rectangle contained by another given line
X, and one segment BC, will be equal to AC2.
Sol.—Erect AD _L to AB, and equal to X. Complete the rect-
angular O ABDE. Construct a square equal to ABDE, and let
AF be one of its sides. Bisect AD.in 6. Join GF. Produce DA
to H. Cut off GEE = GF. In AB take AC = AH. C is the
required point.
Deau—Complete the square AHKC. Produce EC to meet DE
¥66
EXERCISES ON EUCLID.
[book II.
in M. Now DH. HA + AG2 = GH2 (vi.); but GH2 = GF2
= AG2 + AF2; .·. DH . HA = AF2; but AF2 = ABDE (const.);
.·. the figure HM = BD. Reject DC, and HC = BM; but BM
is the rectangle BC. BE ; that is, BC. X; and HC is AC2;
.·. BC. X = AC2.
If we put — = X, where m is any quantity, we get AB . BC
= m AC2.
7. Dem.—Bisect AB in 0. Erect OE A to AB, and join
OC, EP. Now (in., 3) CD is bisected at E;	(x., Ex. 2)
CP2 + PD2 = 2 CE2 + 2 EP2 = 2 CE2 + 2 EO2 + 2 OP* * 2 CO2
+ 2 OP* » 2 AO2 + 2 OP2 =s AP2 + PB2 (ix.).
8. See “ Sequel to Euclid,” Book II., Prop. υπ.
Θ. Let ABCDE be the pentagon; AC, BD, CE, AD, BE its
diagonals. Bisect the diagonals. Let a be the line joining the
middle points of AC, BD ; β of BD, CE; y of CE, AD; 8 of
AD, BE; and · of BE, AC. It is required to prove that 3 (AB2
+ BC2 + CD2 + DE2 + EA2) = AC2 + BD2 + CE2 * AD2 + BE2
+ 4 (a2 + jS* + t2 + 82 + €2).
Dem.—
AB* + BC2 + CD2 + DA2 « AC2 + BD2****2
(xin., Ex. 2).
BC2 + CD2 + DE2 + EB2 = BD2 + CE2 * 4/3T;
CD2 + DE2 + EA2 + AC2CE2 + DA* + 4 72;
DE2 + EA2 + AB2 + BD2 = DA* + EB2 * 4 82 ;
EA2 + AB2 + BC2 + CE2 = EB2 + AC2 * 4
Add together, and we have
3 (AB2 4- BC2 + CD2 + DE2 * EA2) « AC2 * BD2 + CE2
+ AD2* BE2 + 4 (a2 + j82 + t2* 8* + ^).
10.	See “ Sequel to Euclid,” Book II., Prop. v.
11.	See “Sequel to Euclid,” Book IL, Prop. Yin.
12.	See “ Sequel to Euclid,” Book II., Prop. ix.
13.. See “ Sequel to Euclid,” Book Ώ., Prop, at., Cor»BOOK II.]
EXEBCISES ON EUCLID.
67
14.	(1) Dem.—It is proved in Ex. 12 that
m AC2 + n BC2 = m AD2 + n DB2 + (m + n) DC2;
hut wAC2 + n BC2 is given (hyp.); .·. m AD2 + n DB2 + (m +n)
DC2 is given, and mAD2 + » DB2 is given; .·. (w + n) DC2 is
given; but (m + n) is given; .·. DC2 is given; .·. DC is given,
and D is a given point. Hence the locus of the vertex is a Θ,
having D as centre, and DC as radius.
(2) This case can he proved in a similar manner by using
Ex. 13.
15.	Let ABCD he a rectangle, of which AB, AD are adjacent
sides. On AB, AD describe squares AF, AE. Draw the diago-
nals AF, AE. It is required to prove that AF. AE is equal to
twice the rectangle AC.
Dem.—Let fall a X BG on AF. Now, because the L ABF is
right, AF2 = AB2 + BF2 = 2 AB2. For a similar reason AE2
= 2 AD2; hence AF2.AE* « 4AB2.AD2; therefore AF.AE
= 2 AB. AD; that is, AF. AE is equal to twice the rectangle AC.
16.	Dem.—Join AB, BC. Bisect AB in G. Join PG, CG,
AP, BP, CP. Divide GC in H, so that HC = 2 GH. Join PH.
NowAP2 + BP2 = 2 A*G2 + 2GP2 (x., Ex. 2), and 2PG2+PC2
F 268
EXERCISES ON EUCLID.
[BOOK li-
sa 2 GH2 + HC2 + 3 HP2 (Ex. 12); .·. AP2 + BP2 + CP2 = 2 AG2
+ 2 GH2 + HC2 + 3 HP2; but AP2 + BP2 + CP2 is given (hyp.);
.*.2 AG2 + 2 GH2 + HC2 + 3 HP2 is given; but 2 AG2 is given,
and 2GH2, and HC2; hence 3HP2 is given; .·. HP is given,
and the point H is given. Hence the locus of P is a circle.
17. Let ABCD be a square, and AEGH a rectangle of equal
area. It is required to prove that the perimeter of ABCD is less
than that of AEGH.
Dem.—ABCD = AEGH (hyp.). Take away the common part
AEPB, and we have EDCF = BFGH; hence these must be the
complements about the diagonal of a parallelogram; .-.if DC,
AF, HG be produced, they are concurrent. Let them meet in K.
Now DK is greater than DA; . ·. the L DAK is greater than DKA;
that is, CFK is greater than CKF; .·. CK is greater than CF, and
therefore greater than DE. To each add CD + EA, and we get
KD + EA; that is, GE + EA, greater than CD + DA. Hence
the perimeter of the rectangle is greater than that of the square.
18. Let the transversal be divided by the lines, so that m. AC
,, m BC
= n.CB ; then — = —*
λ AC
Dem.—m.AD2 + n.DB2 » #».AC2^ + ». BC2 + (m + n) CD2
(Ex. 12);
,.^AD* + DB*=if AC*+BC* + (; + l)cD*;bu^=g;
+ (gUl)cD-
.·. BC.AD* + AC.DB* = BO.AC* + AC.BC* + AB.CD*;
.·. BC.AD® + AC.DB® - AB.CD® = AC.CB (AC + CB);
.-. BC.AD* + AC.DB* - AB.CD* = AB.B'C.CA.book η.]
EXERCISES ON EUCLID.
69
Lemma.—If a circle be described about an equilateral triangle,
the square of the side of the triangle is equal to three times the
square of the radius.
Dem.—Let BC be the side of the equilateral Δ ABC, and O
the centre of the circumscribing circle. Join BO, and produce
it to meet the circumference in D. Join DC, OC, OA.
Now, in the Δβ AOB, BOO, the L ABO = CBO (I. vm.);
«*. CBO is half an L of an equilateral Δ. In like manner BCO
is half an L of an equilateral Δ, and DOC is an A of an
equilateral Δ, and OD = OC, being radii of the circle; ODC
is an equilateral Δ ; .*. DC = OC; but it has been shown that
BCO is half an L of an equilateral Δ, and DCO an L of an
equilateral Δ ; .·. BCD is a right L; BD2 = BC2 + CD2 = BC2
+ CO2. Let BO be denoted by r, then BD2 = 4r2, and OC2 = r2;
-·. 4r2 ss BC2 + r2. And therefore BC2 = Sr2.
19.	Dem.—Join AD, CD, CD\ Now in the Δ DCD', DD*2
= DC2 + CD#2 + DC. CD' (xn., Ex. 1); .·. 6 DD'2 = 6 DC2
+ 6CD'2+ 6 DC. CD'.
Again, AC2 = 3 CD2 (Lemma), and CB2 = 3 CD'2; .·. AC2. CB2
« 9 CD2. CD'2; .·. AC. CB = 3 CD. CD'; .·. 2 AC. CB = 6 CD. CD';
hence we have 6 DD'2 = 2 AC2 + 2 CB2 + 2 AC . CB = AC2 + CB2
+ (AC2 + CB2 + 2 AC. CB) = AC2 + CB2 + AB2.
20.	—Dem.—Let c be the hypotenuse; then ab—cp (i., Cor. 1);
.\a2b2 = <?p2; .·. a2 b2 = (a2 + b2) p2 = a2p2 + b2 p2. Divide by
f **λ	Ji+i-
21.	Dem.—Since ABD is an isoceles Δ, DC2 — DB2 = AC. CB
<vi., Ex. 6) = AB2 (hyp.). Hence DC2 = DB2 + AB2 = 2 AB2.
22.	Let a variable line AB, whose extremities rest on the cir-
cumferences of two given concentric Θ% subtend a right / at a
fixed point P. It is required to prove that the locus of its middle
point C is a Θ.
Dem.—Join OA, OB, OP. Bisect OP in D. Join CO, CD,
CP.
Now AO2 + OB2 = 2BC2 + 2 CO2 (x., Ex. 2); but AO, OB are
given, being radii of the given Θ8; .*. 2 BC2 + 2 CO2 is given;
BC2 + C02 is given ; but BC = CP (I. xn., Ex. 2); CO2
+ CP2 is given; that is, 2 OD2 + 2 DC2 is given ; but 2 OD2 isEXERCISES ON EUCLID.
[book n.
given, since OP is bisected in D ; .*.2 DC2 is given; .·. DC is a
given line, and D is a fixed point. Hence the locus of C is a Θ,
having D as centre, and DC as radius.BOOK ΙΠ.]
EXERCISES ON EUCLID.
71
BOOK III.
PROPOSITION III.
1. Let AB be the chord subtending a right L at the point C.
It is required to prove that the locus of the middle point of AB
is a circle.
Dem.—Let D be the centre. Draw DE 1 to AB, and join
CD, AD, CE.
A
Now (m.) AB is bisected in E; .·. the lines AE, BE, CE are
equal (I. xn., Ex. 2). Again, AD2 = AE2+ED2 = ED2 +EC2;
but AD3 is given, since AD is the radius; ED2 + EC2 is given,
and the base DC is given; (II. x., Ex. 3), the locus of E is a
circle.
2. Let AB be the given line, and C the given point. Take any
point D in AB. Join DC. With D as centre, and DC as radius,72
EXEBCISES ON EUCLID.
[book m.
describe a Θ. From C let fall a X CE on AB, and produce it
to meet the circumference in F. It is required to prove that
every Θ having its centre in AB, and passing through C, must
pass through F.
Bern.—Take any other point G in AB. Join GO. With G as
centre, and GC as radius, describe a Θ. Join FG. Now EC
= EF (in.), and EG common, and the L CEG = FEG; .·. (I. iv.)
CG = FG. Hence the second circle must pass through F.
3. Let CDE he the given circle, AB the given line, and F the
given point. It is required to draw a chord in CDE which shall
subtend a right L at F, and be || to AB.
Sol.—Let G be the centre of CDE. From G let fall a X GK
on AB. Join FG, and produce it to meet the Θ in E. Bisect
EG in H. Erect HM X to EG, and make it equal to GH.
Join GM. Bisect FG in N, and erect NP X to FG. With G as
centre, and GM as radius, describe a Θ, meeting NP in P.
With N as centre, and NP as radius, describe a Θ, cutting GK
in Q. Through Q draw CD || to AB. CD is the required line.
Dem.—Join GP, GC, CF, QF, QN, FD. Now, since EG
= 2 GH, EG2 =. 4 GH2 ; but MG2 = MH2+HG2=2 GH2. Hence
EG2 = 2MG2 = 2 GP2 = 2PN2 + 2NG2 = 2GN2 + 2NQ2; but
2 GN2 + 2 NQ2 = QG2 + QF2 (II. x., Ex. 2), and EG2 = GC2;
.·. GC2= QG2 + QF2; but GC2 = QC2 + QG2;	QF2 = QC2, and
QF = QC; but QC = QD (in.); hence the three lines QC, QF,
QD are equal; .*. (I. xn., Ex. 2) the L CFD is right.BOOK HI.]
EXERCISES OK EUCLID.
73
PROPOSITION XIII.
1.	(1) Dem.—Let A, B be the centres of the fixed circles, and
P the centre of the variable one. Join AP, BP; and let the
radii be denoted by R, r, r’. Now AP = R+r, and BP = R + r';
.·. AP-BP -r-r\
(2) If the contact of the variable Θ with the Θ whose centre
is B be of the second species, we have AP = R + r, and BP
= R - r'; .·. AP - BP = r + r'.
2.	(1) Dem.—Let the Θ whose centre is P touch that whose
centre is A internally, and be touched by the one whose centre
is B externally; then, denoting the radii as in the last Exercise,
we get AP = r — R, BP = / + R; and .·. AP + BP = r + r\
(2) If the Θ whose centre is B touches the variable Θ in-
ternally, we get AP = r - R, and BP = R - r; .·. AP + BP
= r - r\
3.	Dem.—Let A, B be the centres* and C the point of con-EXERCISES ON EUCLID.
74
[book 111»
tact. Join AB. Through C draw DE, meeting the Θ* in D, E.
Join AD, BE.
Now the L ADC = ACD, and BCE = BEC; but ACD = BCE
(I. xv.); .*. ADC = BEC ; and hence (I. xxvii.) AD is parallel
to BE.
4. Let AB, CD be the diameters, G the point of contact, and
E, F the centres. Join BG. It is required to prove that BG
produced must pass through C.
Dem.—If possible, let it pass through H. Produce DC to
meet BH. Join GE, GF.
Now the L EBG=FHG (I. xxix.); but EBG=EGB = FGH;
.·. FHG = FGH; .·. FG = FH; but FG = FC; .·. FC = FH,
which is absurd. Hence BG produced must pass through C.
In like manner DG produced must pass through A.
PROPOSITION XIV.
1. (1) Dem.—Let ABC be the fixed circle, and AB the chord.
From the centre D let fall a X DE on AB. Join AD.
Now AB is bisected in E (hi.) ; AE is a line of given
length, and AD is given, since it is the radius ; but AD2 = AE2
+ DE2; .·. DE is given, and the point D is given. Hence the
locus of E is a circle.book in.] EXERCISES ON EUCLID.
76·
(2) Let ABC be the Θ, AB the chord, and E any fixed point
in AB.
Bern.—Let D be the centre. Join AD, BD, ED. Now, be-
cause AB is given, and E is a fixed point in it, .·. AE and EB-
are each given; AE . EB is given; and because ADB is an
isosceles Δ, AE. EB = BD3 - DE2 (II. vi., Ex. 6); but AE . EB
is given, and BD2 is given, since BD is the radius; DE is
given, and the point D is given. Hence the locus of E is a circle.
PROPOSITION XV.
1. Let ABC be the Θ, and P the point. Through P draw a
chord AB ± to the diameter CD. It is required to prove that AB·
is the minimum chord.
Dem.—Through P draw any other chord EG; and from E,
the centre, let fall a A EH on it. Now the Z EHP is right,
EPH is acute; ΈΡ is greater than EH; .·. (xv.) FG is
greater than AB.
2. Let ABC be the given circle, AB the given chord, and P
the point. It is required, through the point P, to draw a chord
equal in length to AB.76
EXERCISES ON EUCLID.
[BOOK Π2.
Sol.—From the centre D let fall a _L DE on AB. With D as
centre, and DE as radius, describe a Θ EFG. Through *P draw
PCFH, touching EFG in F, and cutting ABC in C and H.
CH is the chord required.
Dem.—Join DF. Now because DF=DE, .·. (xiy.) CH=AB.
3.	See “ Sequel to Euclid,” Book III., Prop. xy.
4.	Dem.—Let O, O', 0" he the centres. Now the lines
joining 00', ΟΌ", O'O must pass through A, C, B (in.).
And because OA = OB, the L OBA = OAB. Similarly, the L
O"BD = 0"DB; hut 0"BD » OBA; hence 0"DB = OAB, and
0"D is parallel to OA. In like manner 0"E is parallel to
O'A; and hence 0"D, 0"E are in the same straight line.
PROPOSITION XVI.
1. Dem.—Let D he the common centre, and AB, CH the
•chords of the greater which touch the less; then AB = CH
•(xiy.). See diagram to Prop, xv., Ex. 2,BOOK III.]
EXERCISES ON EUCLID.
77'
2.	Let AB be the given line, and 0 the centre of the given
circle. It is required to draw a parallel to AB which shall touch
the circle. (See last diagram.)
Sol.—Let fall a _L OC on AB; and through D, where OC’
cuts the Θ, draw EF parallel to AB. EF is the required line.
Dem.—Now the L ODF = OCB (I. xxix.); .*. ODF is a
right L; hence (xvi.) EF touches the circle.
3.	Let AB be the given line, and O the centre of the given
circle. It is required to draw a perpendicular to AB which shall
touch the circle.
Sol.—From 0 let fall a _L OC on AB. Draw OF || to AB,
and from F, where it meets the Θ, draw FB (| to OC. FB is
the required line.
Dem.—The Z.8 OCB, FBC are together equal to two right L8
(I. xxix.) ; .·. the L FBC is right, and FB is ± to AB.
4.	(1) Sol.—Let O be the given point, and AB the given line.
Let fall a X OC on AB. With 0 as centre, and OC as radius,
describe a circle.
(2) Let 0 be the given point, and O' the centre of the given
circle. It is required to describe a Θ having its centre at 0, and
touching the Θ whose centre is O'.
Sol.—Join 00', and produce to meet the circumference of O
in C; with 0 as centre, and OC as radius, describe a Θ; or,
with 0 as centre, and OD as radius, describe a Θ. Hence there
are two solutions.
5.	Let AB, AC be the given lines, and R the given radius.
It is required to describe a Θ, touching AB, AC, and having a
radius equal to R.
Sol.—Erect AD 1 to AB, and equal to R. Draw DE || to AB.78	EXERCISES ON EUCLID. [boox hi.
Bisect the L BAC by AE, meeting the line DE in E. E is the
-centre of the required 0.
Dem.—Draw EB, EC, A 8 to AB, AC.
Now the L BAE = CAE, and the right Z.· ABE, ACE are
equal, and AE common; (I. xxvi.) BE = CE ; and the Θ,
with E as centre and BE as radius, will pass through C. If we
produce BA, we can describe another 0 touching AC and the
production of BA.
6.	Let AB, AC be the given lines, and E the centre of one of
the Θ* which touch AB, AC.
Sol.—Join AE, and produce it. Join E to the points B, C,
where the Θ touches AB, AC. Now, since the Z.· at B, C are
right (xvi.), AE2 = AB* + BE2 = AC2 + CE2 ; but BE2 = CE2 ;
.·. AB2 = AC2 *, AB = AC, AE common, and the base BE
= CE ;	(I. vin.) the Z. BAE = CAE, .·. the L between the
lines is bisected by the line joining their intersection to the centre
of one of the circles. Hence the locus of the centres is the right
line bisecting the angle between the two given lines.
7.	(1) Let C be the centre of the given circle, AB the given
line, and R the radius* It is required to describe a 0 that shallBOOS ΙΠ.]
EXERCISES ON EUCLID.
79
touch the Θ whose centre is C and the line AB, and have a
radius equal to R.
Sol.—Take any point A in AB, and erect AD 1 to it and = R;
draw DE || to AB; from C draw any radius CF, and produce it
to G, so that FG = R. With C as centre, and CG as radius,
describe a Θ cutting DE in E. E is the centre of the required
circle.
Bern.—Join CE, and draw EB || to AD. Now CG = CE,
and CF «= CH; .·. FG = EH ; but FG = R; EH = R, and
EB = AD = R, EH = EB, and the Θ, with E as centre and
£B as radius, will pass through H. Hence it will touch the
given Θ, the given line, and have a radius of given length.
(2) Let C, C' be the centres of the given Θ8, and R the given
radius.
SoL—Draw any two radii CD, C'D', and produce them to
E, E', so that DE, D'E' are each equal to R; with C, C' as
centres, and CE, C'E' as radii, describe two Θ8. Let them in-
tersect in C". C” is the centre of the required circle.
Dem.—Join CC", C'C". Now CE = CC", and CD CF ;
hence DE = FC"; but DE = R (const.); .·. FC" = R. In like
manner F'C" = R; .*. the Θ described with C" as centre, and
C"F as radius, will pass through F', and touch the two Θ8, and
have the given radius.
PROPOSITION XVII.
2. Let 0 be the common centre. From any points A, B, on
the outer circle tangents AC, BD are drawn to the inner one. It
is required to prove that AC « BD.80
EXERCISES ON EUCLID.
[book m.
Dem.—Join OA, OB, OC, OD. Now (xvi.) the at C, D
are right; .·. OA2 = OC2 + CA2, and OB2 = OD3 + DB2; but
OA2 = OB2, and OC2 = OD2, .·. AC3 = BD2; .·. AC = BD.
3.	Let ABCD be the quadrilateral. It is required to prove
that AB + CD = AD + BC.
Dem.—Let E, F, G, H be the points of contact. Now (xvn.,
Ex. 1) AE = AH, and BE = BF ; .·. AB = AH + BF. In Bhe
manner CD = DH + CF; .·. AB + CD = AD + BC.
4.	Dem.—Let ABCD be the circumscribed parallelogram.
Now AB + CD = 2 CD, and AD + BC = 2 AD ; but AB + CD
= AD + BC; .*. 2 CD = 2 AD ; .·. CD = AD. In like manner
all the sides are equal. Hence ABCD is a lozenge.
Again, the line joining the centre to the intersection of tan-
gents bisects the angle between the tangents; conversely, the
line bisecting the angle between the tangents passes through the
centre; therefore AC passes through the centre. Similarly, BD
passes through the centre. Hence E is the centre.
5.	Dem.—OB = OD, and OP common, and the base BP = DP;
(I. vm.) the L BOP = DOP. Again, OB = OD, OF com-
mon, and the L BOF = DOF; .·. (I. iv.) the L OFB = OFD.
Hence each is a right Z., and OP is JL to BD.
6.	Let A, B be the centres of the Θ8. Let P be a point from
which the tangents PC, PD to the Θ* are equal. It is required
to prove that the locus of P is a right line.
Dem.—Join AC, AP, BD, BP, and from P let fall a X PE on
AB. Now AP2 = AC3 + CP3; .·. CP2 = AP3 - AC3. In like
manner DP* = BP2- BD2; but CP2 =. DP3; .·. AP3 — AC* = BP2
-BD2; AP2 - BP2 = AC3- BD3; but AC2 - BD8 is given,
since AC, BD are the radii of the Θ"; .*. AP2 - BP3 is given,*oox m.]
EXEBCISES ON EUCLID.
81
/. AE2 — EBa is given; .% E is a given point; hence EP ia given
in position, and therefore the loons of P is the right line EP
(called the radical axis of the two circles).
Cor.—To construct the line EP, join the centres, divide the
joining line in E, so that AE2 — EB2 = AC2 — BD2; and erect
EP 1 to AB.
7.	Let the three circles be denoted by A, B, C. It is required
D
to find a point such that the tangents from it to A, B, C shall be
equal.
Sol.—Find a line DE, such that the tangents from any point
of it to A and B will be equal (xvn., Ex. 6) ; and find a line FE,
such that the tangents from any point of it to A and C shall be
equal. E, where the lines DE, FE intersect, is the required
point.
8.	Dem.—OBP is a right-angled Δ, and BF is 1 to OP (xvn.,
Ex. 5); .·. (I. xlvii., Ex. 1) OB2 = OF . OP.
9.	Let AB, AC be two fixed tangents, and EF a variable tan-
gent, cutting AB, AC in E, F, and touching the Θ in D. Let
0 be the centre. Join OE, OF. It is required to prove that the
L EOF is constant.
Bern.—Join OB, OC, OD. Now (I. vm.) the L EOD = EOB ;
EOD = JBOD. In like manner FOD = J COD; .·. EOF
= \ BOC; but the L BOC is constant, since the tangents AB,
AC are fixed; .·. the L EOF is constant.
10.	(1) See “ Sequel to Euclid,” Book III., Prop. vxi.
G82	EXERCISES ON EUCLID. [book in.
(2) Draw a line cutting two circles, X, Y, so that the inter-
cepted chords shall be of given lengths A, B.
Sol.—Let 0, 0' be the centres of X, Y; R, R' their radii.
Then with 0, 0' as centres, describe Θ· CDE, FGH, the squares
A
of whose radii shall be equal to R2 — | A2, and R'2 — J B2 respec-
tively, and draw the line IM a common tangent to both circles.
IM is the line required.
Bern.—Let C, F be the points of contact. Join OC, 01;
O'F, O'L. Now OC2 = OI2 - IC2 = R2 - IC2; but OC2 = R2
— \ A2 (const.); .*. IC2 = J A2. Hence IC = £ A ; but IC
= J IK (III. in.); .*. IK =* A. In like manner LM = B.
PROPOSITION XXI.
1. (1) Let ABC be a Δ, whose base BC, and vertical LBAC,
are given. From B, C let fall _L· BE, CF on AC, AB, and let
them intersect in G. It is required to find the locus of G.
Bern.—The four L8 A, F, G, E of the quadrilateral AFGE
are together equal to four right L ■ (I. xxxii., Cor. 3); but the
L* E, F are right; .*. the Z.8 A, G are together equal to two
right Z8; but A is given (hyp.); .*. G is given; .·. (I. xv.)
the L BGC is given. And hence (xxi., Cor. 2), the locus of G
is a circle.
(2) Let the internal bisectors meet in D. Now, the three L8
of the Δ ABC are equal to two right L 8; but the L A is given;
*. the sum of the L8 B, C is given; half their sum is given;BOOK ΙΠ.]
EXERCISES ON EUCLID.
83
that is, DBC + DCB is given; the L BDC is given; and
hence (xxi., Cor. 2) the locus of D is a circle.
(3)	Let the external bisectors meet in D. Then, as before, the
sum of the L* B, C is given; .·. (I. xxxn., Ex. 14) the L D
is given. Hence (xxi., Cor. 2) the locus of D is a circle.
(4)	Dem.—Let the external bisector of the L C, and the in-
ternal bisector of B meet in D; then the L BDC = } BAC
(I. xxxn., Ex. 2); the L BDC is given. Hence (xxi.,
Cor. 2) the locus of D is a circle.
2. Let AB2 be equal to the sum of the squares of the two lines.
It is required to prove that their sum is a maximum when the
lines are equal.
Sol.—Upon AB describe a semicircle ADB. Bisect AB in C,
and erect CD _]_ to AB. Join AD, BD. In ADB take any other
point E. Join AE, BE. Produce AD to F, so that DF = DB.
Join BF. Produce AE to G, so that EG = EB, and join BG.
Dem.—The L DFB = DBF (I. v.); but BDF is a right L;
..·. DFB is half a right L. Similarly, EGB is half a right L;
hence (xxi., Cor. 1) the four points A, F, G, B are concyclic.
Now, since D is a point in a Θ from which the three equal lines
DA, DB, DF are drawn to the circumference, D is the centre;
AF is the diameter; but the diameter is the greatest chord;
.*. AF is greater than AG ; that is, the sum of AD rana DB is
greater than the sum of AE and EB.
3. Let there be two Δ® ADB, AEB on the same base AB, and
having equal vertical Z.8, and let ADB be isosceles. It is re-
quired to prove that the sum of the sides AD and DB is greater
•than the sum of the sides AE and EB. (Diagram, Ex. 2.)
<*284
EXERCISES ON EUCLID.
[book m.
Dem.—Produce AD to F, so that DF = DB. Join BF. Pro-
duce AE to G, so that EG = EB, and join BG. Now the, L DFB
= DBF (I. y.) ; hut ADB = DFB + DBF (I. xxxn.) ; .·. ADB
= 2 DFB. Similarly, AEB = 2 EGB ; hut ADB = AEB (hyp.);
.·. DFB = EGB ; and .*. (xxi., Cor. 1) the points A, F, G, B·
are concyclic; and it can be shown, as in Exercise 2, that AD
4- DB is greater than AE + EB.
4. Dem.—Let ABC he an inscribed Δ. Then if any two sidea
AC, CB he unequal, by supposing the points A, B to remain fixed
while C varies, the perimeter will he increased by making AC, CB
equal. Hence, when the three sides AB, BC, CA become all equal,,
the perimeter will he a maximum.
Lemma.—Let ABC, DBC he twoA8 on the same base inscribed
in a circle, of which ABC is isosceles. It is required to prove that
the area of ABC is greater than the area of BDC.
Dem.—Through A draw AF, touching the Θ. Produce BD
to meet it in F, and join CF. Now the L FAC = ABC {xxxn.)
= ACB; .·. AF is J to BC; hence (I. xxxvn.) the Δ BFC
BAC ; but BFC is greater than BDC; BAC is greater than
BDC. Similarly it can be shown that BAC is greater than any
other Δ inscribed in the Θ, having BC for base, whose sides are
unequal. Hence the area of the isosceles Δ is a maximum.
5. Let ABCDE be a polygon inscribed in a circle. It is re-*
quired to prove that the area is a maximum when all the sides are
equal.
Dem.—Join AC. Now, if we suppose the point B to move
about whilst the others remain fixed, when AB — BC, the Δ ABC
will be a maximum (Ex. 5), and therefore the area of the wholeBOOK ΙΠ.]
EXERCISES ON EUCLID.
85
figure will be increased. In like manner, if any other of the
sides be unequal, we can increase the area by making them
equal. Hence the area will be a maximum when all the sides are
equal.
PROPOSITION XXII.
1.	Let ABCD be a quadrilateral, whose opposite L* B, D are
supplemental. It is required to prove that it is cyclic.
Bern.—If not, let the Θ through A, D, C, intersect the line
AB produced in E. Join CE. Now the L8 ADC, CBA are
together equal to two right L8 (hyp.), and the L* ADC, CEA
are equal to two right L8 (xxn.) Reject ADC, and we have
the L CBA = CEA, which is impossible (I. xvi.). Hence the
Θ must pass through B.
2.	Let ABCDEF be a hexagon inscribed in a Θ. It is re-
quired to prove that the sum of the alternate L8 ABC, CDE, EFA
is equal to four right angles.
Bern.—Join CF. Now the L8 ABC, CFA are together equal
to two right L8 (xxn.), and the L8 CDE, EFC are equal to two
right L8. Hence, by addition, the L8 ABC, CDE, EFA are
-equal to four right angles.
3.	(1) Let ABDC be a cyclic quadrilateral, and let the opposite
sides meet in E, F. Draw any line GK, cutting the four sides,
and making the L EJX = EKJ. It is required to prove that the
ZGHF = HGF.
Bern.—The L* BDC and BAC are equal to two right L*
«(xxn.), and the L8 BAC, BAG equal to two right L8. Reject the86
EXERCISES ON EUCLID.
[book m*.
L BAC, and we have the L BDC = BAG, and the L DKJ
= AJG (hyp.); .·. the remaining L DHK = AGJ; that is, the
L GHF = HGF.
(2) Let GK cut the diagonals in Μ, N. It is required to prove
the L OMN = ONM.
Bern.—The L EJK = EKJ (hyp.), and the L ABC = ADC
(xxi.); the remaining L BMJ = DNK; that is, the L OMN
= ONM.
4.	Bisect the L AEC by ES, meeting the diagonals in Q, R.
From 0 let fall a X OP on ES. It is required to prove that OF
bisects the L QOR.
Dem.—The L ABC = BER + BRE (I. xxxii.), and ADC
= DEQ + DQE; but (xxi.) ABC = ADC; .·. BER + BRE
= DEQ + DQE; but BER = DEQ (hyp.); BRE = DQE
= OQR, and the L OPR = OPQ. Hence the L ROP = QOP.
5.	Let ABCDEF be a cyclic hexagon, having the side AB || to-
DE, and BC to EF. It is required to prove that the side AF i&
1| to CD.
Dem.—Join CF. Now the L ABC = DEF (I. xxix.,
Ex. 8) ; and since ABCF is a cyclic quadrilateral, the L8 ABC,
AFC are together equal to two right L8. For the same reason
the L* DCF, DEF are equal to two right L%; .·. the L* ABC
and AFC = DCF and DEF; but ABC = DEF; .·. AFC = DCF.
And hence (I. xxvii.) AF is || to CD.
6.	Dem.—Join AB. Now the L BAD = BFD (xxi.), and
BAC = BEC; .*. BFD = BEC. And hence (I. xxviii.) CE m
|| to DF.BOOK ΙΠ.]
EXERCISES ON EUCLID.
87
7. On the sides of any Δ ABC, equilateral Δ* are described,
BF and CD joined and intersecting in G. Join AG, EG. It is
required to prove that AG and GE are in the same straight
line.
Dem.—Since AB = AD, and AC = AF, and the L BAD
= CAF; to each add BAC, therefore the L DAC = BAF ; hence
(I. iv.) the L ADC = ABF, and ACD = AFB. Now, because
the L ACG = AFG, AFCG is a cyclic quadrilateral; hence
the L 8 AFC, AGO are together equal to two right L8
(xxii.); similarly ADBG is a cyclic quadrilateral; and the L 8
ADB, AGB are equal to two right L8; .·. these four L8 are
together equal to four right L % and the L9 AGB, BGC, AGC
are equal to four right L*. Reject the L* AGB, AGC, and we
have the L BGC equal to the sum of AFC and ADB. To
each add BEC, and we have BGC + BEC = AFC + ADB
·+- BEC; hut these three L8 are equal to two right L8, since each
is an L of an equilateral Δ ; BGC, BEC are equal to two
right L8; and hence BGCE is a cyclic quadrilateral; the L
EGC = EBC; EGC is equal to an L of an equilateral Δ, and
therefore equal to AFC; hut AFC and AGC are equal to two right
L%\ EGC and AGC are equal to two right Z.8, and hence
(I. xrv.) AG and EG are in the same straight line. Therefore
AE, BF, CD are concurrent.
8. If we join the centres H, J, K, it is required to prove that
HJE is an equilateral triangle.88
EXERCISES ON EUCLID.
[book hi.
Dexn.—Let HI, JK, HE cut BG, CG, AG in the points
L, Μ, N. Now, because the L· L, N are right (ni., Cor. 4),
.·. the L% H, G are equal to two right L·, and the L· G, D
are equal to two right L8; hence the L H = D; .*. H is an L
of an equilateral Δ. Similarly, K is an L of an equilateral A .
Hence the A HJK is equilateral.
9.	Let ABCD be the quadrilateral, 0 the centre of the inscribed
circle, and E, F, G, H the points of contact. Join 0 to A, B, C, D.
It is required to prove that the L ■ [AOB, DOC are supple-
mental.
Dem.—Join OE, OF, OG, OH. Now the L AOB = half sum
of the L · EOH, EOF (xvn., Ex. 9), and the L DOC = half sum
of GOH, GOF; but the sum of EOH, EOF, GOH, GOF is four
right L8; .*. AOB and DOC are together equal to two right
angles.
10.	Let ABC be a A, whose perpendiculars CD, BE intersect
in G. Join AG, and produce it to meet BC in F. It is required
to prove that AF is 1 to BC.
Bern.—Join DE. Now, because each of the L8 ADG, AEG
is right, ADGE is a cyclic quadrilateral; hence the L DEG
= DAG (xxi.) Again, since the L* BDC, BEC are right, the
points B, D, E, C are concyclic, and therefore the L DEB = DCB;
.·. DAG = DCB, and DGA = FGC (I. xv.); .·. ADG = AFC ; but
ADG is. a right L; AFC is a right Z., and AF is perpen-
dicular to BC.
11.	Let a variable tangent CD meet two parallel tangents
AC, BD. Join the centre 0 to C, D. It is required to prove
that the L DOC is right.book m.]
EXERCISES ON EUCLID.
89
Bern.—Draw the diameter AB, and join 0 to the point Έ
where CD touches the circle.
Now the L DOC is equal to half the sum of the L · EOB,
EOA (xvn., Ex. 9); but EOB and EOA are together equal to
two right L* ; .·. the L DOC is right.
12.	See “ Sequel to Euclid,.” Book III., Prop. xn.
13.	Let ABCDEF be the hexagon, 0 the centre of the inscribed
•circle, and Ο, H, J, K, L, M the points of contact of the hexagon
and circle. Join 0 to the points A, B, C, D, E, F. It is required
to prove that the sum of the L 8 AOB, COD, EOF is two right
angles.
Dem.—Join 0 to the points G, H, J, K, L, M. Now the L
AOB = JMOH (xvn., Ex. 9), COD = §HOK, and EOF
= J KOM; .-.the sum of AOB, COD, EOF is two right L·.
1. Let AB, CD he the two diameters given in position. Take
any point E in the circumference, and let fall _L8 EF, EG· on
AB, CD. Join FG. It is required to prove that FG is given in
magnitude.
Dem.—Join OE, and from A let fall a X AH on CD. Now,
since the L OHA is right, the circle on OA as diameter will pass
through H (xxxi.); and because the L8 OFE, OGE are right,
the 0 on OE as diameter will pass through F and G; but OA
= OE; .*. the Θ8 on OA and OE are equal, and the L AOH
is in both those Θ8; the arc AH is equal to the arc FG
(xxvi.), and therefore the chord AH = FG; but AH is given in
magnitude, since it is a X from the extremity of one of the
PROPOSITION XXVIII.EXERCISES ON EUCLID.
90
[book in*
diameters given in position on the other. Hence EG is given
in magnitude.
2. Let OA, OD be two lines given in position, and FG a line
of given length sliding between them. At the extremities of FG,
perpendiculars EF, EG are erected to OA, OD. It is required
to prove that the locus of E, where these perpendiculars meet,
is a circle. (Diagram to Ex. 1).
Dem.—Join OE. Erect ON _!_ to OD, and equal to FG ;
draw NA || to OD.
Now, because ONA is a right L, the Θ described on OA as
diameter will pass through N ; for a similar reason, the Θ on OE
as diameter will pass through F and G. Now since ON and FG
are equal, and subtend equal angles OAN, FOG in the Θ8 OAN,
FOG, the Θ8 are equal; therefore the diameters OA, OE are
equal. Again, since ON = FG, ON is given, and AN is || to
OD ; .·. the point A is given, and hence the line OA is given in
magnitude; but OE = OA; .*. OE is given in magnitude, and
the point 0 is given. Hence the locus of E is a Θ, having 0 as
centre and OE as radius.
PROPOSITION XXX.
1. Dem.—Let 0 be the centre. Through C draw CG H to
DA. Join OB, OC, OG. Now the L GCO = COE (I. xxix.);
hut GCO = CGO, and CGO = AOG, .·. DOC = AOG; .·. the arcbook m.] EXERCISES ON EUCLID.	9l·
DC = AG (xxvi.). Again, the L GOfe is double GCB (xx.)
but GCB = AEB (I. xxix.), and AEB = COE; because CE
as OC (hyp.); .·. GOB is double DOC; hence the arc GB is
double CD, and therefore the arc AB is three times the arc CD.
2. (1) Let AD be the internal bisector of the vertical L of
the Δ ABC. Join BD, CD. It is required to prove that BD*
= CD.
Dem.—Because the L BAD = CAD, the arc BD = CD (xxvi.),
and therefore the chord BD = CD (xxix.).
(2) Produce CA to E. Bisect the L BAE by AF, meeting the
circumference in F. It is required to prove that the point F is
equally distant from B and C.
Dem.—Join BF, CF. Now the L8 FBC and FAC are together
equal to two right L* (xxn.), and FAC and FAE are equal
to two right Z.8 (I. xiii.) ; .*. the L FBC = FAE. Again,
the L ?AF = BCF (xxi.); but BAF = FAE, BCF = FAE;
BCF = FBC ; .·. BF = CF.
3. Dem.—The L ADB = AD'B (xxi.), and the L ACB = AC'B;
but the L 8 ACB and DCB are together equal to two right L8,92
EXERCISES ON EUCLID.
[book m.
and the L8 AC'B, D'C'B are together equal to two right L*;
.·. the L DCB as D'C'B ; and hence (I. xxxii., Con. 2) the
remaining Ls DBC, D'BC’, are equal.
4. Bern.—Join AD, DB. Now because the L ACD = BCD,
the line AD = BD. Again, the L 8 DBC, DAC, together eqnal to
two right L* (xxii.), and the L8 DBC, DBF equal to two right
L8 (I. xiii.); the L DAE = DBF, and the right L8 DEA, DFB
are equal; .·. (I. xxvi.) AE = BF. Hence AC — CE = CF
- CB ; .·. AC + CB = CF + CE = 2 CE; because CF = CE.
1. Let the circles touch in A. Through A draw any line BAC.
It is required to prove that BAC divides the circles into similar
segments.
Dem.—Through A draw a common tangent DE; take any
D
PROPOSITION XXXII.
D

points F, G, in the Θ8. Join AF, BF, AG, CG. Now the L BAD
= AFB (xxxii.), and the L CAE = AGC; but BAD = CAE (I. xv.);book πι.] EXERCISES ON ETTCIID.	95-
AFB = A6C; and hence the segments AFB, AGC ace-
similar.
2.	Let the circles touch in A. Through A draw two lines BC,
FG, meeting the Θβ in B, C; F, G. Join BF, CG. It is
required to prove that BF, CG are parallel.
Bern.—Through A draw a common tangent DE. Now it
may be proved, as in Ex. 1, that the L AFB = AGC; hence
(I. xxvii.), BF is parallel to CG.
3.	Dem.—Join AC, BC. Now the lines CA, CD, CE are
D
equal (I. xh., Ex. 2); .·. the L AEC = EAC ; but (xxxn.) EAC
= BCE) hence the L BCE = BEC) .*. BCE and BEC — 2 BEC j
(I. xxxn.) the L CBA = 2 BEC ; but BEC = CAB, since CE
= CA; .·. CBA = 2 CAB. Hence the arc AC = 2 CB.
4.	(1) See “ Sequel to Euclid,” Book III., Prop. in.
(2) Dem.—Let GBF and ECH be the tangents to the circles
at the points B, C. Join CF, CG. Now the L GFC = FCH
(I. xxix.); but FCH = FGC (xxxii.) ; GFC = FGC ; and
hence the chords GC, FC are equal.
5.	(1) Let the circles ABC, DBE touch at B. Draw a com-
mon tangent AD. Join AB, DB. It is required to prove that
the angle ABD is right.
Dem.—Draw a common tangent BF. Now AF = BF (xvn.r
(Ex. 1); .*. the L ABF = BAF; and because BF = DF, the
Z.BDF = DBF; .·. the L ABD = BAD -t- BDA, and hence
(I. xxxn., Cor. 7) the L ABD is right.
(2) Dem.—Produce AB, DB to meet the circumferences in
E, C. Join AC, DE. Produce ED to G, and draw AG parallel
to CD.
Now, because the L ABD is right, EBD is right, and therefore
ED is a diameter, and hence (xix.) the L ADE is right; .*. AD94	EXERCISES ON EUCLID.	[book m.
is 1 to EG. Again, since AG, CD are parallel, the L GAE
= DUE, and is therefore a right L. Hence (I.Jxlvii., Ex. 2)
AD2 = DE. DG = DE. AC.
PROPOSITION XXXIII.
1.(1) Let AB be the base, X the vertical L, and P the per-
pendicular.
Sol.—On AB describe a segment ACB containing an L equal
-to X (xxxiii.). Erect AD X to AB, and = P. Through D draw
DC || to AB, cutting the circle in C. Join AC, CB. ACB is the
required triangle.
Dem.—Let fall a X CE on AB. The vertical L ACB = X;
AB is the base, and the X CE = AD = P.
(2) Let the sum of the sides be equal to S.
Sol.—On AB describe a segment ACB containing an L equal
J X. Produce AB to E. Cut off AE = S. With A as centre, and
AE as radius, describe a Θ, cutting ACB in C. Join AC, BC,
and at the point B in the line BC make the L FBC = FCB.
AFB is the required triangle.book m.]
EXERCISES ON EUCLID.
96
Dem.—FC = FB (I. τι.); .·. AC = AF + FB; but AC = AE
— S; .·. AF + FB=S; and theZ.AFB:= FBC + FCB (I. xzxu.)
= 2 FCB = X.
(2*) Let MN be the base, D the difference of sides, and ABC
the vertical angle.
1 Sol.—Produce AB to E. Bisect the L CBE by BF. On
MN describe a segment MGN containing an L = ABF; in MN
take MH =D. With M as centre, and MH as radius, describe
a Θ, cutting MGN in G. Join MG, NG. Produce MG, and at
the point N in GN make the L GNK = NGK. MEN is the
required triangle.
Dem.—Produce MK to L, and draw KP parallel to GN.
Now KN = KG (I. vi.); MG is the difference between MK
and NK; but MG = MH = D; .*. the difference between MK
and NK is equal to D. Again, the L PKN = GNK (I. xxix.),
and LKP = KGN; but GNK and KGN are equal (const.); .·. PKN
and LKP are equal; and since the L MKP = MGN = ABF,EXERCISES ON EUCLID.
96
[book m.
the L LKF = EBF ; but LKP = NKP, and EBF = FBC, .·. FBC
= NKP. Hence the L MKN = ABC.
(3) Let AB be the given base, X the vertical L, and let the
sum of the squares of the sides be equal to 2 S*.
Sol.—On AB describe a segment containing an L — X. Bisect
AB in D, and erect DE ± to AB; from A inflect AF on DE = S
(I. ii., Ex. 2). With D as centre, and DF as radius, describe
a Θ, cutting ACB in C. Join AC, BC. ACB is the Δ re-
quired.
Dem.—Join CD. Now, DF = DC ; .·. DF2 = DC2; AD2
+ DF» = AD2 + DC2; .*. AF2, that is S2 = AD2 + DC2 ; but
AC2 + CB2 = 2 AD2 + 2 DC2 (II. x., Ex. 2). Hence AC2 + CBg
= 2 S2.
(3') Let AB be the base, X the vertical L, and D2 the difference
of the squares of the sides.
Sol.—On AB describe a segment ACB containing an L — X.
Divide AB in E, so that AE2 - EB2 = D2 (“Sequel,” Book I.,
Prop. ix.). Erect EC 1 to AB, and join AC, BC. ACB is the
triangle required.BOOK ΙΠ.]
EXERCISES ON EUCLID.
9t
Dem.-AC* = AE2 + EC2, and BC2 = BE2 + EC*; . . AO2
— BC* = AE* — EB2 = D2.
(4) Let AB be the base, X the vertical L, and S the side of
the inscribed square.
Sol.—On AB describe a segment containing an L = X. Erect
AD 1 to AB. In AD take AE = S. On AE describe a square
AEFG. Join BF, and produce it to meet AD in H. Through
H draw HC [| to AB, meeting the Θ in C. Join AC, BC.
ACB is the required triangle.
Dem.—Produce EF to meet AC, BC in J, K; and draw JL,
KM || to AE. Now, JK = EF (I. xxxviii., Ex. 6); but EF
= AE = JL; .·. JK = JL ; hence the sides of JKLM are equal,
and the L 8 are right (const.) ; .*. it is a square, and is inscribed
in the Δ ABC.
(5) Let AB be the base, M the median, and X the vertical
angle.
Sol.—On AB describe a segment ACB containing an L = X ;
bisect AB in D. With D as centre, and a radius equal to M,
describe a Θ, cutting ACB in C. Join AC, BC, DC. ACB is
the triaqgle required.
Dem.—Because D is the centre of the Θ cutting ACB, DC is
the radius ; but the radius is equal to M; .*. DC = M, and it is
the median bisecting the base AB.
H98
EXERCISES ON EUCLID.
[book m~
2. Let A be the fixed point, and 0 the centre of the given
circle. Take any point B in the circumference of the Θ. Join
AB, and bisect it in C. It is required to prove that the locus of'
C is a circle.
Dem.—Join AO, OB, and through C draw CD || to OB.
Now AO is bisected in D (I. xl., Ex. 3) ; but A and O are
given points, .·. the point D is given; and since CD is parallel to-
OB, CD = J OB; but OB is a given line; CD is given, and
the point D is given. Hence the locus of C is a Θ, having D as
centre and DC as radius.
3. Let AB be the base, and ACB the vertical L. About ACB”
describe a segment of a circle containing an L — ACB ; then the-
circle must pass through C. On AC, BC describe equilateral Δ*
ADC, BEC. Join DE. It is required to find the locus of the
middle point of DE.
Dem.—On AB describe an equilateral Δ AFB. Join CF, DFr
EF. Now the L BAF = DAC, .*. the L BAC = DAF; and
since DA = AC, and BA = AF, we have DA and AF equal AC
and AB, and the contained L8 are equal; hence (I. iv.) DP
= CB — CE. Similarly, DC = EF; .·. DCEF is a parallelogram;
hence (I. xxxrv., Ex. 1) DE, CF bisect each other in G. Now
F is a given point, and C a point on the circumference of thebook m.]
EXERCISES ON EUCLID.
99
circle, and FC is bisected in G; (xxxm., Ex. 2) the locus of
G is a circle.
4. Let ACB be a Δ, whose base and vertical L are given. On
DC describe a square BEDC. It is required to find the locus of
E. On AB describe a square ABGF. Join EG. Now AB and
BC = GB and BE, and the contained L* are equal; (I. iv.)
the L ACB = BEG; BEG is a given L, and the base BG is
given, since it is equal to AB; .*. (xxi., Cor. 2) the locus of E
is a circle.
PROPOSITION XXXV.
1. Let ACB be the triangle. About ACB describe a circle.
Draw the diameter CE, and from C let fall a JL CD on AB. It
is required to prove that AC. CB = CD . CE.
Dem.—Join AE. Now the L CAE is right (xxxi.), and is
equal to CDB, and the L AEC = ABC (xxi.); .*. (I. xxxn.,
Cor. 2) the L ACE = BCD, and hence (xxxv., Cor. 3) AC. CB
* CD. CE.
2. Let ABD he a circle, of which AC is the diameter; let AB
be the chord of an arc, then BC is the chord of its supplement.
Join B to the centre E. Let fall a A BF on AC, and .produce it
h2EXERCISES ON EUCLID.
100
[book in.
to meet the circumference in D. It is required to prore that
AB.BC = BE.BD.
Dem.—Join CD. Now the L BDC = BAC (xxi.); but BAC
=* ABE, and BDC = DBC) .*. ABE = DBC | hence the Δ· ABE,
DBC are equiangular; and .·. (xxxv., Cor. 3) AB. BC = BE.BD.
3. Let ABC be a triangle whose base and the sum of whose
aides are given. Produce AC to D, and bisect the L BCD by EF.
From A, B let fall _L» AF, BE on EF. It is required to prove
that AF . BE is given.
Dem.—Produce BE to meet AD. Bisect AB in G. Join EG,
FG. Now because the L BCE = DCE, and CEB = CED, each
being right, and CE common, .·. (I xxvi.) BE = DE, and BC
= DC. Now, since BC = DC, .·. AD = AC + CB; hence AD
is given; and because AB, DB are bisected in G, E, .·. GE is
| to AD, and equal to half AD (I. xl., Exs. 2 and 5); that is,
= i (AC + CB). Similarly, GF = J(AC + CB); .·. the Θ, with
G as centre, and GE as radius, will pass through F, and will be
a given Θ. Produce EG to meet the circumference in H, and
join AH. Now because AG = GB, and GH = GE, and thebook m.]
EXERCISES ON EUCLID.
101
L AGH = BGE, (I. rv.) AH * BE, and the L GAH = GBE.
To each add the L GAF, and we have the L * GAH, GAF
= GBE, GAF; but GBE, GAF are equal to two right L*f since
BE and AF are parallel, .*. GAH and GAF are equal to two right
L *; hence AH, AF are in the same straight line. Now FGH is
an isosceles Δ, (II. τι., Ex. 6) AF . AH = FG2 — AG*; but
FG is given, since it is half the sum of AC and CB; and AG is
given, because it is half AB. Hence AF. AH s given; that is,
AF. BE is given.
4. Let ABC be a triangle whose base AB, and the difference of
whose sides AC, CB is given. Bisect the L ACB by CD. From
A, B let fall the A· AD, BE on CD. It is required to prove
that AD . BE is given.
Dem.—Produce BE to meet AC in F. Bisect AB in G. Join
EG, and produce it to meet AD in H. Join GD. Now because
the L BCE = FCE, and the L BEC = FEC, and CE common;
(I. xxvi.) CB = CF, and EB = EF, .*. AF is the difference
between AC and BC; and because EB = EF, and GB = GA, GE
is | to AF, and equal to half AF (I. xl., Exs. 2 and 5) or
half EH; .·. GE = GH; and the three lines HG, EG, DG are
equal (I. xii., Ex. 2); .*. the Δ HGD is isosceles; hence
(II. vi., Ex. 6) AD . AH = AG2 — GH2; but AG is given,
since it is half AB; and GH is given, because it is equal EG
= J AF; .·. AD . AH is given; that is, AD. BE is given.
5. Let ACD, BCD be two circles intersecting in C, D. At D
draw a tangent to the Θ BCD, meeting ACD in E. From G, the
centre of ACD, let fall a A GH on DE, and let it meet ACD inEXERCISES ON EUCLID.
102
[book III.
A. Join AD, and produce it to meet BCD in B. AB is the
required line.
Bern.—Draw AK || to DE. Join CD, and produce it to meet
AK. Take any other point L in ACD, and draw LM || to DE.
Join LD, and produce it to meet BCD in F. Join CF, CB. Now
the L EDC = CBD (xxxii.) ; hut EDC = AKC (I. xxix.), AKC
= CBD, .·. AKBC is a cyclic quadrilateral; hence vxxxv., Cor. 3)
AD. DB = CD . DK. In like manner LMFC is a cyclic quadri-
lateral ; .·. LD . DF = CD. DM ; but CD . DK is greater than
CD . DM, AD . DB is greater than LD . DF.
8. Let AB, AC he two lines given in position, and P a given
point. It is required through P to draw a transversal, so that
PE. PB = S2.
Sol.—Join AP, and produce it to D, so that AP.PD = S2.
On PD describe a segment of a Θ PED, cutting AC in E, and
containing an L — BAD. Join ED, EP, and produce EP to
meet AB. EPB is the required line.«οοκ m.]
EXERCISES ON EUCLID.
103
Dem.—Because the L PED = BAD, AEDF is a cyclic quadri-
lateral ; .·. (xxxv., Cor. 3)EP.PB = AP.PD; but AP.PD is
•equal to S2. Hence EP. PB is equal to S2.
PROPOSITION XXXVI.
1. Dem.—Describe a circle about the triangle ACB; then
because the L BAD = ACB, AD is a tangent (xxxii.). Hence
<xxxvi.) DB.DC = DA-.
PROPOSITION XXXVII.
1. (1) Let A, B be the given points, and EC the given lirne^
It is required to describe a Θ passing through A, B, and touching
the line EC.
Sol.—Join AB, and produce it to meet EC. Find a point D
in EC, so that CD2 = AC . CB ; and through the points A, B, D
describe a circle. Then because CD2 = AC. CB, the line CE
touches the circle.
(2) Let A, B be the given points, and CDE the given circle.
It is required to describe a Θ passing through A, B, and touch-
ing CDE.
Sol.—Describe a Θ passing through A, B, and cutting CDE
in D, E. Join AB, DE, and produce them to meet in F. From
F draw a tangent FC to CDE, and through the points A, B, C
describe a circle. ABC is the required circle.
Dem.—DF.FE = AF.FB; but DF. FE = FC2; .·. AF.FB104
EXERCISES ON EUCLID,
[book m*.
=d?C2; .*. CF touches ABC, and it (ouches CDE. Hence ABCT
touches CDE, and it passes through A and B.
2. (1) Let AB, AC be the given lines, and P the point. It is
required to describe a Θ, touching A B, AC, and passing through P.
Sol.—Bisect the L BAC by AD. From P let fall a X PE on
and produce it until EF = EP; let it meet AB in B. InAR»
take a point G, so that BG2 = FB . BP. Erect GH X to AB; and’
from H, where GH meets AD, let fall a X on AC. The Θ
through the points F, P, G will be the required circle.
Bern.—Because BG2 = FB. BP, the Θ passing through F, P, G
must touch AB ; and since AB touches the Θ, and GH isXto AB,
GH passes through the centre (xix.), and AD passes through the
centre (hi.) ; .*. H is the centre, and (I. xxvi.) theA8 AGH, AKH
are equal in every respect; HG = HK ; but HG is the radius
··. HK is the radius. Hence the circle must touch AC in K,
3. Let AB be the line, CDE the Θ, and P the point.
Sol.—From P let fall a perpendicular PF on AB, and produce-
it until F G *= PF; and through P and G describe a circle PEG,,
touching CDE (Ex. 1). PEG is the required circle.BOOK OT.]
EXERCISES ON EUCLID.
105<
Dem.—Because PG is bisected at right L' by AB, the centre of
4. Let ABCD be the given Θ, P, P' the points.
Sol.—Draw a line AB, cutting off an arc AB in ABCD equal
to the given arc* Let E be the centre* From E draw EF _L to
AB. With E as centre, and EF as radius, describe a Θ FG.
Through P, P' describe a Θ PP'HK, cutting ABCD in Η, K.
Join EH, PP', and produce them to meet in L. Through L
draw LCGD, a tangent to FG, and cutting ABCD in C, D. The
Θ through P, F, C will be the required one*
Dem.—Join EG. Now because PP'KH and DCKH are cyclic
quadrilaterals, PL. LP' = HL. LK = DL. LC; hence PP'CD is a
cyclic qtialrilateral; the Θ through P, P', C must pass through
D; and since E is the centre of FG, EF = EG; .·. AB = CD (xiv.),
and therefore the arc AB = CD. Hence through P, P' we have
described a Θ PP'CD, intercepting an arc CD = AB, on a given
O ABCD.106
EXERCISES ON EUCLID.
[book III.
5. Dem.—Let 0, O' be the centres. Join OE, O'F. Now
since OE, O'F are each X to EF, they are || to each other; hence
the L DOE = DO'F; but the L BOE is (III. xx.) double of the
___E
A	Ο	B C O' D
L BAE, and DO'F is double of DCF; hence the L BAE =
DCF. In like manner, the L ABE = CDF. Hence the Δ5
ABE, CDF are equiangular.
6. If r be the radius of the inscribed circle of a right-angled
triangle, by making the construction, we see at once that 2r is
equal to the excess of the sum of the legs above the hypotenuse.
Again, if p, p' be the radii of circles touching the hypotenuse,
the X from the right angle on the hypotenuse, and the Θ described
about the right-angled Δ, it follows at once from the Demonstra-
tion, Book VI., Ex. 59, that p + p* is equal to the same excess.
Hence 2 r — p-v pr.
Miscellaneous Exercises on Book III.
1. Let AB, CD be two chords of a circle intersecting at right
L8. It is required to prove that the sum of the squares of the
four segments is equal to the square of the diameter.
Dem.—Draw BF || to CD. Join CB, FD, AF, AD. Now
<JB2 = CE2 + EB2; but CB = FD (xxvi., Cor. 2); .-.FD2 = CE2
+ EB2, and AD2 = AE2 + ED2 ; .*. AD2 + FD2 = AE2 + EB2
+ CE2 + ED2; but since the L ABF = AED (I. xxix.), .·. ABF
is a right L ; hence AF is the diameter; .*. the L ADF is right;
.·. AF2 = AD2 + DF2 = AE2 + EB2 + CE2 + ED2.
2. (1) Let AB, a chord of a given Θ, subtend a right L at aBOOK III.]
EXERCISES ON EUCLID.
107
fixed point P. From P, and C, the centre of the Θ, let fall Xs
PE, CD on AB. It is required to prove that CD. PE is constant.
A
Dem.—Join CP, CA, PD, and let fall a X PQ on CD. Now
AB is bisected in D (hi.) ; the lines AD, DP, DB are equal
<1. xii., Ex. 2), and AC2 = AD2 + DC2 = DC2 + DP2; but DC2
-f DP2 is greater than CP2 by 2 CD . DQ, (II. xm.); that is, by
2 CD . PE; .·. AC2 is greater than CP2 by 2 CD . PE; but AC2
and CP2 are given; .·. CD. PE is given.
(2) Join CP, and bisect it in F. Erect FG X to CP, and equal
to CF or PF. Produce GF to H, so that FH = FG, and join
•CG, PG, CH, PH. From C, G, Η, P let fall X8 CJ, GD, HK,
PE on AB. It is required to prove that GD2 + HK2 is constant.
Dem.—Because CGPH is a square, GD2 + HK2 is greater than
2 CJ. PE, by the area of CGPH (“ Sequel/’ Book II., Prop, vm.);
but 2 CJ. PE is given (1), and the area of the square is given.
Hence GD2 + HK2 is given.
3. Let the Θ· intersect in A, B. Through B draw a line BCD,
meeting the Θ8 in C, D. Join AC, AD. It is required to prove
that AC = AD.
Dem.—Because the Θ8 are equal, the arcs AB are equal;EXERCISES ON EUCLID.
108
[book III.
.·. the L· ACB, ADB are supplemental. Hence the L· ACDr
ADC are equal. And hence AC = AD.
4. (1) Let AB, AC be two fixed tangents, and EF a tangent
cutting off with AB, AC an isosceles Δ AEF. AEF is greater
than any other Δ AHO, made by a tangent HG, which does not
cut off an isosceles Δ with AB, AC.
Bern.—Let EF, HG intersect in J. Join OJ, OB, OC, OD,
OG, OH, and produce OH to L. Now, because AB = AC, and
AE = AF; .·. BE = CF; but BE » DE, and CF = DF; DE
=s DF; JF is greater than JE.
A
Again, the L HOG = BOD, because each = £ BOC (xvn.),
Ex. 9), and HOJ « £ BOD ; .·. HOJ = JOG, and the L HJO
= KJO, and JO common; (I. xxvi.) JH = JK. Now
the L LHG is greater than HGO; but LHG = GKJ, because
they are the supplements of the equal L* OHG, OKG ; GKJ
is greater than JGK ; JG is greater than JK; JG is greater
than JH, and JF is greater than JE; .·. the Δ FJG is greater
than EJH. To each add the figure AGJE, and we have the
Δ AEF greater than AHG.
(2) Let the tangent be drawn below the Θ, making an isosceles
Δ with the fixed tangents ; then it can be shown, as in (1), that
the isosceles Δ is less than the Δ formed by any other tangent
which does not cut off an isosceles Δ with the fixed tangents.
5. Dem.—Join CF, DE, AB. Now the Z* ADE and ABEBOOK III.]
EXERCISES ON EUCLID.
109
are equal (xxi.), and ACF, ABF equal; .·. ADE, ACF are equal;
CE is || to DF;	CDEF is a parallelogram, and.·. (I. xxxiv.)
CD = EF.
6.	See Book I., Miscellaneous Ex. 45.
7.	Let the sides of the cyclic quadrilateral ABCD be the dia-
meters of four circles. It is required to prove that those circles
intersect in four concyclic points E, F, G, H.
Dem.—Draw the diagonals AC, BD, and let fall JL* AE, BF,
CG, DH on AC, BD. Join EF, EH, GF. Now, because the
L · AHD, CHD are right, the Θ· on AD, CD, as diameters, will
pass through H. In like manner the Θβ on the other sides will
pass through E, F, G. And eince the L · AHD, AED are right,
AHED is a cyclic quadrilateral; .·. the LB AHE, ADE are to-
gether equal to two right L9 (xxn.), and the L · AHE, FHE are
equal to two light L ■; .·. the L ADE = FHE. Similarly, BCF
*= EGF; but ADE = BCF (xxi.); .·. FHE = EGF. And hence
(xxi., Cor. 1) the points E, F, G, H are concyclic.
&8. Let ABCD be a cyclic quadrilateral. Draw the diagonals110
EXERCISES ON EUCLID.
[book ΠΙ.
AC, BD. It is required to prove that the orthocentres of the
Δ· ADB, ACB, CAD, CBD are the angular points of,a quadri-
lateral which is equal to ABCD.
Dem.—From D and C let fall ±8 DE, CF on AB. Let C', D'.
he the orthocentres of the Δ8 ADB, ACB; and let A', B' be the
orthocentres of the Δ8 BCD, ADC. Join C'D', D'A\ A'B', B'C';
and from 0, the centre, let fall a _L OM on AB.
Now OM = £CD' (“Sequel,” Book I., Prop, xn., Cor. 3).
Similarly OM = J C'D; .·. CD' = C'D, and they are parallel;
hence DCD'C' is a parallelogram, .·. DC = D'C7. In a similar
manner it can he shown that the other sides of A'B'C'D' are
respectively equal and parallel to the remaining sides of ABCD.
Hence A'B'C'D' = ABCD.
9.	Let the circles intersect in A, B. Through A draw ACD,
AEF, cutting the Θ8 in C, E ; D, F. Join EC, FD, and produce
them to meet in G. It is required to prove that EGF is a given
angle.
Dem.—Join AB, BC, BD. Now the L8 BAE, BCE are equal
to two right L8 (xxn.), and BCE, BCG are equal to two right L8
(I. xm.); BAE = BCG. Similarly BAE = BDG, .*. BCG =
BDG, and hence (xxi., Cor. 1) the points B, C, D, G are concyclic;
.·. the L CBD = CGD. Again, the L8 ACB, ADB are given,
since they are in given segments, and the L CBD is equal to
ACB — CBD; .·. CBD is a given L; that is, CGD is a given
angle.
10.	See “Sequel to Euclid,” Book III., Prop. x.
11.	Let P, P' be the points in the circle.
Sol.—Join PP'. Bisect it in D. Join D to the centre E, and
produce it to meet the circumference in C. C is the point
required.KOOK III.]
EXERCISES ON EUCLID.
Ill:
Take any other point B in the circumference. Join BP, BP\
CP, CP', BD. Now because E is the centre, DC is greater than
DB, .*. 2 DC2 is greater than 2 DB2. To each add 2 DP2, and
we have 2 DC2 + 2 DP2 greater than 2 DB2 + 2 DP2; but CP2
+ CP'2 = 2 DC2 + 2 DP2 (II. x., Ex. 2), and BP2 + BP'2 = 2 DB*
+ 2 DP2; .·. CP2 + CP'2 is greater than BP2 + BP'2. Hence CP*
+ CP'2 is a maximum. In like manner it can be shown, if we
produce CD to A, that AP2 + AP'2 is a minimum.
12.	Let ABCD (see fig., Ex. 7) be the quadrilateral. Draw
AC one of the diagonals; and from B, D let fall 1* BF, DH
on AC. It is evident, from the proof of Ex. 7, that BF and
DH are the common chords of the Os on CD, AD, and on AB,
CB as diameters, and that they are parallel.
13.	See “ Sequel to Euclid,” Book III., Prop. xt.
14.	Let ACB be the Δ, and CD the internal bisector of tho
vertical L. It is required to prove that AC. CB = CD2
+ AD. DB.
Dem.—Describe a 0 about ACB. Produce CD to meet the
circumference in E, and join BE. Now the L ACE = BCE, and
CAD = CEB (xxi.); .·. (I. xxxii., Cor. 2) the Δ8 ACD, BCE
are equiangular; hence (xxxv. Cor. 3) AC . CB = EC . CD; but
EC.CD = ED. DC + CD2 (II. m.), and ED. DC = AD.DB
(xxxv.), AC. CB = CD2 + AD. DB.
15.	Draw BD, CD tangents to the circles. It is required to
prove that BDC is a given angle.
Denr.—Join AE, BE, CE. Now the L DCE = CAE (xxxii.),
and DBE = CAE; .·. DCE = DBE, and CFD = BFE (I. xv.),
CDF = BEF; but BEF = ABE — ACE (I. xxxii.), and112	EXERCISES ON EUCLID. [book in.
ABE and ACE are given L8; .·. the L BEF, that is, CDF, is
given.
16. Let AB, a chord of a given circle, pass through a given
point P; at A, B tangents AD, BD are drawn. It is required to
prove that the locus of D is a right line.
Dem.—Let E be the centre. Join ED, EP. Produce
EP, and from D draw DF JL to it. Now, denoting the
radius by R, we have (xvn., Ex. 8) DE. EG = R*; but because
the L8 DGP, DFP are right, DFPG is a cyclic quadrilateral,
and.·. DE.EG = FE.EP; .·. FE.EP = R2; .*. FE.EP is
given, and EP is given; EF is given; hence F is a given
point, and FD is X to EF; .·. FD is a line given in position.
Hence the locus of D is a right line.
17. Let ABC he the triangle. Describe a Θ about ABC.
Bisect the L BAC by AJ, and produce it to meet the circum-KOOK III.]
BXEBCISES OK EUCLID.
113
ference in D. Through D draw the diameter DE. From. A let
fall a X AL on BC. Produce AC to G ; and let fall 1· DF, DO
on AB, AG; then CG = J(AB - AG) (Dem. of xxx., Ex. 4).
It is required to prove that HJ.HL = CG*.
Dem.—Join FH, GH, DC, CE, EA, and from A let fall a
£
X AM on DE. Now the L EAD is right (xxxi.), and EHJ is right,
EAJH is a cyclic quadrilateral, ED.DH = AD.DJ; but
because the L ECD is right, and CH X to ED, ED. DH = DC*
(I. xlvii., Ex. 1); .·. AD.DJ = DC*, and AD.DK = DG-;
hence, by subtraction, AD. JK = CG2; and since the Δ· ADM,
HJK are equiangular, we have (xxxv., Cor. 3) AD. JK = HJ. AM
= HJ. HL. Hence HJ. HL = CG2.
18. The rectangle contained by the distances of the point where
the external bisector of the vertical L meets the base, and the
point where the X from the vertex meets it from the middle point
of the base, is equal to the square of half the sum of the sides.
Let the same construction be made as in Ex. 17. Join EA,
and produce it to meet BC produced in N; then EA is the ex-
ternal bisector of the vertical L (xxx., Ex. 2). It is required
to prove HN. HL = AG2*
Dem.—Through H draw HO || to AD, meeting EN in 0, and
AM in P. Now the L * NOH, AMD are equal, each being right,
and the L PAJ * PHJ (I. xxxiv.); .·. the L MDA = ANH,
the Δ· HNO, AMD are equiangular, .·. (xxxv., Cor. 3)
HN. AM = DA.OH; but AM = HL, and OH^AK, .·. HN. HL
= DA . AK ; but (I. xlvii., Ex. 1) DA . AK a* AG*. Hence
HN. HL = AG*.114
EXERCISES ON EUCLID.
[book III.
19. Let ABCD be a cyclic quadrilateral. Produce AB, DC to
meet in E, and AD, BC to meet in F. Join EF; and from E, F
draw tangents t, f to the 0 described about ABCD. It is re-
quired to prove that EF2 = i2 +
Dem.—About the Δ CDF describe a 0 CDFG, cutting EF
in G. Join CG. Now (xxn.) the L 8 BAD, BCD are together
equal to two right L8, and the L8 DFG, DCG are equal to two
right L*; .·. the 1} BAD, BCD, DFG, DCG are equal to four
right L 8; and the L 8 BCD, BCG, DCG are equal to four right
Ls. Reject BCD, DCG, and we have the L BCG » BAD -f DFG.
To each add the L BEG, and we get BCG 4- BEG = EAF + AFE
+ AEF; hence the L8 ECG, BEG are equal to two right L8;
.·. BCGE is a cyclic quadrilateral; FE.EG = DE.EC
= *2 (xxxvi.), and EF . FG = BF. FC = i'2 ; but EF2 = FE. EG
+ EF.FG; .·. EF2 = t2 + t'2.
20. Let AB be a given line, CDE a given O, and DKJ a
variable 0, touching CDE in D, and AB in J. It is required to
prove that JD produced passes through a given point.
Dem.—From the centre F let fall a X FH on AB, and produce
ittomeetthe 0 in C. Let G be the centreof DKJ. Join FG, GJ,
CD, and produce JG to L. Now (xx.) the L LGD = 2GJD
= 2 GDJ, and the L EFD = 2 FDC; but LGD = EFD (L xxix.),
.·. GDJ = FDC, .·. JD, and DC are in one straight line; that
is, the chord of contact JD produced passes through the point Cbook m.]
EXERCISES ON EUCLID,
115
where the X from the centre of the giyen Θ on the given line
meets the circumference.
C
21. Dem.—Join DA, AE. Now the L DEA = DAB (xxvii.),
and EAC * ADE; but AFG = FDA + FAD (I. xxxu.) and AGF,
» GAE + GEA, .·. AFG = AGF, and hence (I vi.) the lines AF
=and AG are equal.
22. Dem.—Join BD, B'D', and produce them to meet in F.
12.EXERCISES ON EUCLID.
116
[book
Join. EC, E'C',V and produce them to meet in G. Produce
FB', GE to meet in F', and FB, GE' to meet in G'.
Now the Z BDE = BCE (xxi.), and B'D'E' = B'C'R'; hut
B'D'E' = DD'F (I. xv.), and B'C'E# = CC'G; hence the Z DFD''
= CGC', and .·. (xxi. Cor. l)the four points F, G, F', G' are
concyclic.
23. Let ABCD be"a cyclic quadrilateral, such that a circle can
he inscribed in it. It is required to prove that the lines A'B',
C'D', joining the points of contact, are perpendicular to each
other.
Dem.—Because AC' and BD' are tangents, if we produce-
them until they meet, they will be equal, the Z AC'D'
= BD'C'. To each add the Z CD'C', and we have AC'D' + CD'C'
= BD'C' + CD'C'; but BD'C' + CD'C' equal two right Z8;
.*. AC'D' + CD'C' equal two right Z *. Similarly, AA'B'+ CB'A"
equal two right Z % and (xxn.) DAB + DCB equal two right Z8,
.·. the sum of those six Z" is six right Z *; and those Z ®, to-
gether with the L * A'OC' + B'OD' equal eight right Z8
.·. A'OC' + B'OD equal two right Z8; but A'OC' = B'OD'.
Hence each is right, and therefore A'B' and C'D' are perpen-
dicular to each other.
24. Let ABCD be a cyclic quadrilateral; AC, BD its diagonals
intersecting in E. Through E draw the minimum chord FG
(xv., Ex. 1). It is required to prove that EH = EJ.
Dem.—Through C draw CK || to FG, and join KE, KH, KD.
Now, because FG is bisected in E, and CK'is | to FG, .*. ECfl»OOK III.]
EXERCISES ON EUCLID
117
= BS) and the L JEC = HEK; hut TEC — ECKj ·*« TTRX
= ECK; but ECK = ADK (xxi.); .·. HEK = ADK, and
A
_·. HEDK is a cyclic quadrilateral; the L HDE == HKE;
but HDE = ACB (xxi.); .·. HKE = ACB. And the Δ® EHK,
EJC have two L · and a side in one equaleto two L * and a side in
the other. Hence (I. xxvi.) EH = EK.
25.	See “Sequel to Euclid,’* Book VI., Sec. i., Prop. xv. (3).
26.	See “ Sequel to Euclid,” Book III., Prop, xx., Cor. 2.
27.	Let AB, AC, BD, CE be four,lines forming four Δ· ABD,
ACE, BEF, DCF. About the Δ® BEF, DCF two Θ» are de-
scribed intersecting in F, G. It is required to prove that the
Θ· about the Δ* ABD, ACE will pass through G.
Dem.—Join GB, GF, GD. Now the L BEF = BAC -f ACE;
.but ACE = FGD (xxi.); BEF = BAC + FGD; .·. BEF118
EXERCISES ON EUCLID,
[book m~
+ BGF *= BAD + BGD; but (xxn.) BEF + BGF equal two
right Z8; .·. BAD + BGD equal two right Z8; hence the Θ
about BAD will pass through G. Similarly the Θ about ACE
will pass through G.
28.	About AYZ, CXY describe Θ8 intersecting in 0. It »
required to prove that the Θ about BXZ will pass through O.
Dem.—Join OX, OY, OZ. Now the Z8 ZAY + ZOY equal
two right Z8 (xxn.), and YCX + YOX equal two right Z8;
.·. those four Z3 equal four right Z8, and the three Z8 ZOY,
YOX, XOZ equal four right Z8; hence the Z XOZ = ZAY
+ YCX ; ZOX + ZBX - BAC + ACB + CBA, and .·. equal
two right Zs. Hence the Θ about ZBX will pass through O.
29.	Dem.—Join OC, OB. Now, because the points O and G
are given, the line OC is given in position, and YC is given iu
position ; .·. the Z YCO is given; (xxi.) the Z YXO is given.
In like manner OXZ is given; hence the Z YXZ is given.
Similarly, it can be shown that the L9 XZY and XYZ are
each given.
30.	Let XYZ be a given Δ, and A, B, C three given points.
It is required to place a Δ equal to XYZ whose sides shall pass
through A, B, C.
Sol.—Join AB, AC, BC. On BC, AC describe segments con-
taining L8 respectively equal to the Z8 X, Y. Join 0, O', the·
centres. On 00' describe a semicircle, and in it place a chord.
0*D = \ XY. Through C draw AB (I to O'D. Join BA,DEPARTMENT OF MATHEMATIC?
COBMUUNIVERSiry ι,α
HOOK ra.]	EXERCISES ON EUCLID.	119
A'B, and produce them to meet in CV A'B'C' ie the required
triangle.
Bern.—From O' let fall a ± O'F on A’B’. Join OD, andpro-
duce it to meet A'B' in E. Now the L 0D07 is right (xxxi·).
OEF is right; hence (hi.) B'C is bisected in E, and CA' is
bisected in F; .·. B'A' == 2 EF = 2 O'D = XY; and since the Z·
A', B' = X, Y respectively, the Δ A'B'C' = XYZ.
31. Let XYZ be the given Δ, and AB, AC, BC the given lines.
It is required to place a Δ equal to XYZ whose vertices shall be
on AB, AC, BC.
Sol.—Through the points X, Y, Z describe a Δ ΧΎ'Ζ' equal
to ABC (Ex. 30), and . in BC take BA' = Y'Z; in BA take BC*
= Y'X, and in AC take B'C = ΧΎ. Join A'B', B'C', C'A'.
A'B'C' is the Δ required.120
EXERCISES ON EUCLID.
[book III.
Dem.—Because A'B = Y'Z, and BC' = XT', and the L A'B'C'
= XY'Z ; .·. (I. it.) A'C = XZ. Similarly A'K = YZ, apd B'C'
= XY. Hence the Δ A'B'C' = XYZ.
32. Let ABC be the given Δ, and X, Y, Z the three points.
It is required to construct the greatest Δ equiangular to ABC,
whose sides shall pass through X, Y, Z.
Sol.—Join XZ, YZ, and on them describe segments of Θ8 con-
taining Z* respectively equal to the Z® B, C. Through Z draw
BO' |j to the line joining the centres. Join B'X, C'Y, and pro-
duce them to meet in A'. A'B'C' is the Δ required.
Dem.—Through Z draw any other line DE. Join DX, EY,
and produce them to meet in F. Now (xxi.) the Z EDF
= C'B'A', and the Z DEF = B'C'A', and the side B'C' greater
than DE (“ Sequel,” Book III., Prop. xv.). Hence the Δ A'B'C'
is greater than DEF.
33. Let AB, AC, BC be the three given lines, and A'B'C' the
given Δ. It is required to construct the minimum Δ equi-
angular to A'B'C, whose vertices shall be on AB, AC, BC.
Sol.—On BC describe a segment of a Θ containing an Z equal
to the sum of the Z® A, A'. On AB describe a segment contain-
ing an L equal to the sum of the Z* C, C'. From D let fall
Is DE, DF, DG on AB, AC, BC. Join EF, GF, EG. EFG
is the required triangle.
Dem.—The Z CDB = A + A' (const.); but CDB = A + DCF
+ DBE; ,\A' = DCF + DBE. Again (const.), FCGD and EBGDhook m.]
121
EXERCISES ON EUCLID.
;are cyclic quadrilaterals; .*. the L FCD = FGD, and DBE = DOB;
hence the L FGE = FCD + DBE; hence the L FGE = A'. Simi-
larly GFE = B', and GEF = C'. Therefore the Δ FGE is equi-
angular to A'B'C'.
Draw any line DG', and draw DF', DE', making each of the
La FDF', EDE' equal to GDG'. Join G'F', F'E', E'G'. Now
the L FDF' = GDG'. To each add FDG', and we have the
L F'DG' = FDG. To each add the L F'CG, and we get F'DG'
+ F'CG' = FDG + FCG; hut FDG + FCG = two right l*;
.-. F'DG' + F'CG' = two right Z.*; hence F'CGD is a cyclic
quadrilateral; the L F'G'D = FCD ; but FCD has been shown
to be equal to FGD;	. F'G'D = FGD. Similarly E'G'D = EGD;
.·. F'G'E' = FGE. In like manner G'F'E' = GFE, and F'E'G'
= FEG. Hence the Δ* F'E'G', FEG are equiangular; and since
DG' is greater than DG, and DF' greater than DF, and the
L G'DF' = GDF, the side G'F' is greater than GF; .‘.the
Δ F'E'G' is greater than FEG. Hence FEG is a minimum.
34. Let 0, O', 0" be the centres of the given Θ8. It is re-
quired to construct the greatest Δ equiangular to a given one,
whose sides shall touch the three circles.
Sol.—Through the points 0, 0' 0", describe the maximum
Δ ABC, equiangular to the given one. Draw tangents A'B',
B'C', C'A' respectively || to AB, BC, CA. A'B'C' is the re-
quired Δ.
Bern. — From A, B, C let fall _]_■ on the sides of the
Δ A'B'C'. Because the L * about B are together equal to four122
EXERCISES ON EUCLID.
[book in-
right Z8; and that the L8 EBA, E'BC are each right, the
L8 EBE', ABC are together equal to two right Z8; hut ABC
is a given L ; EBE' is given, and the sides BE, BE' are
given, since they are equal to the radii of the Θ8 0, 0". Hence
the figure EBE'B' is given in magnitude. Similarly the figures
ADA'D', CFC'F' are given in magnitude! Again, since the
Δ ABC is a maximum, the side BC is a maximum; therefore
BCFE' must be a maximum, because it is contained by BC and
BE', which is a given line, being equal to the radius of 0".
In like manner each of the figures ABDE, ACD'F' is a maximum-
Hence the whole figure A'B'C' is a maximum.
35. Let AB, AC, two sides of a given Δ ABC, pass through
tw o fixed points P, P'. It is required to prove that the side BC
touches a fixed circle.
Dem.—Join PP'. Describe a Θ about the Δ APP\ Draw
the diameter AD, and join DP, DP'. From D let fall a A DE on
BC, and produce it to meet the Θ in F. Join AF, and let fall
a 1 AG on BC.
Now since the points P, P' are given, PP' is a given line, and
the Z PAP' is given; hence (xxi., Cor. 2) the circle PAP' is
given; and because the Z8 EBP, EDP are together equal to two
light Z8, and EDP, FDP are together two right Z8, .·. the
Z FDP = EBP, and is therefore a given L; hence the arc PF ia
given, and .*. F is a given point. Again (xxxi.) the Z AFD is
right, and FEG is right; hence AFEG is a parallelogram; EF
= AG ; but AG is given, since it is the A from the vertex on the-
baee of a given Δ ; .*. EF is given, and the point F is given;123-
book m.] EXERCISES ON EUCLID,
hence the locus of E is a Θ, haying F as a centre, and EF as
radius. Hence the base BC touches a fixed circle.
36. Let the sides CA, CB of the Δ ABC touch fixed Θ8. It is
required to prove that AB touches a fixed Θ.
Dem.—Through the centres 0, 0' draw ||· A'C7, B'C' to AC,
BC. Join 0, O' to the points of contact D, E; and through
A, B, C draw ||* AF, BG, CH, CJ to OD, O'E.
Now the L BAC = B'A'C'; .·. BA'C' is given, and the L AFA'
is right; .*. the Δ AA'F is given in species, and the side AF is
given, being equal to OD; .·. AA', A'F are each given. Again,,
the L9 ACB, HCJ are equal to two right L8; but ACB is given
HCJ is given, and the sides CH, CJ are given; HCJC' is a*
given figure; C'H is given, and HF is given, being equal to
AC; .·. A'C is a given line. Similarly B'C' is given, and A'B'
is given; the Δ A'B'C' is given. And hence (Ex. '35 A'B'
touches a fixed circle.
37. Let P be the given point, and C the centre'of the circle.124
EXERCISES ON EUCLID.
[book III.
Sol.—Join PC, and on it describe an equilateral Δ PAC.
Draw PB, bisecting the L APC. From B let fall a 1 BD on
PC, and produce it to meet the Θ in E. Join EP. EPB is the
required triangle.
Dem.—BD = ED (in.), and DP common, and the L BDP
= EDP; .·. (I. iv.) PB = PE, and the L BPD = EPD; but
A
BPD is \ an L of an equilateralΔ ; .·. EPD is J an L of an equi-
lateral Δ. Hence EPB is an L of an equilateral Δ* and the
L PEB = PBE. Hence the Δ EPB is equilateral.
38. Let P be the given point, and C, Cr the centres of the
given Θ*. It is required to construct an equilateral Δ, having
its vertex at P, and the extremities of its base on the circum-
ferences of C and C\
Sol.—Join PC, and on it describe an equilateral Δ PAC.
With A as centre, and a radius equal to CD, describe a Θ, cut-kook in.]
EXERCISES ON EUCLID.
125.
ting the Θ whose centre is C' in B. Join AB, and at the point
C in CP make the L PCE = BAP (I. xxiii.). Join BE, EP,
PB. BEP is the required Δ.
Dem.—Because AP — CP, and AB = CE, and the L BAP
= ECP, .·. (I. iv.) the base BP = EP, and the L BPA = CPE.
To each add the L APE,	the angle BPE — CPA, hence BPE is
an L of an equilateral Δ. And since PB = PE, the Δ PBE
is equilateral.
39. Let ABC be a given Δ. It is required to place it so that
its sides shall touch three given Θβ 0, O', 0".
Sol.—If two sides of a Δ equal to ABC touch two Os O, 0',.
the third must touch a fixed Θ (Ex. 36). Let PP' be the fixed Θ.
Draw DE a common tangent to 0", PP' (xvii., Ex. 10).
Through 0, 0' draw OE', O'D', meeting DE produced, and
making the Z.8 OE'D', O'D'E' respectively equal to the Z8 CBA,
CAB. At 0, 0' draw OF', OG' at right Z.8 to OE', O'D'; and
through F', G draw EF, DF, touching the Θ*. DEF is the
Δ required.
Dem.—Because each of the Z8 EOF', EF'O is right, E'O,
EF are ||; .*. the L DEF = DE'O, and .·. equal CBA. Simi-
larly, EfDF is = CAB. Hence DEF is the Δ required.
40. Let ABCD be a given quadrilateral. It is required to-
describe a square about it.126
EXERCISES ON EUCLID.
[book nr.
Sol.—On AD, BC, two opposite sides, as diameters, describe
Θ· AED, BFC. Bisect the semicireles A ED, BFC in E, F.
Join EF, and produce to meet the Θ· again in G, H. Join HB,
GA, and produce them to meet in J. Join GD, HC, and produce
them to meet in X. GJHK is the required square.
Dem.—Because the arc AE = DE, the L AGE = DGE; but
the L AGD is right (xxxi.), .*. AGE is half a right L . In like
manner BHF is half a right L, .*. AGE = BHF, .·. JH = JG.
Similarly, KG = KH; hence the sides are equal, and the L 6 are
evidently right. Therefore GJHK is a square.
Lemma.—To find a point 0 in a Δ ABC, such that the L BOO
A
may exceed the L BAC by a given L X, and that the L AOC
may exceed the L ABC by a given L Y.
Sol.—On BC describe a segment of a Θ containing an L equal
to BAC + X, and on AC describe a segment containing an L equalHOOK III.]
EXEBCISES ON EUCLID.
127
to ABC + T. The point 0, in which these segment» intersect, is
•evidently the required one.
41. Let ABCD be a given quadrilateral. It is required txy
inscribe a square in it.
Sol.—Produce AB, DC to meet in E, and AD, BC to meet in
F.	In the Δ AED find a point 0, such that the L AOD is equal
to AED, together with a right L , and that the L DOE is equal
to FAE, together with half a right L (Lemma); and in AFB find
a point O', so that the L AO'B is equal to AFB, together with a
right L ; and the L AO'F = ABF, together with half a right L .
Describe a Θ through the points 0, O', A; cutting AF, AE in
G,	H. Through 0, G, D describe a Θ, cutting DE in J; and
through O', Η, B describe a Θ, cutting BF in I. Join GJ, JI,
IH, HG. GJIH is the required square.
Dem.—Join OG, OH, OJ. Now the difference between the
L8 AOD and AED is equal to a right L (const.), and AOD
-AED = EAO + ODE; hence EAO and ODE are together
equal to a right L; but EAO = HGO (xxi.), and ODE = OGJ,
hence the L HGJ is right. Similarly, by joining JH, it can be
shown that GJH is half a right L, GH = GJ. Similarly, it
can be shown that the L GJI is right, and that GJ = JI. Hence
GJIH is a square.
42. (1) Lemma.—To find the radical axis of a Θ and a point.
Let A be the centre of the Θ, and B the point.to the square of the radius AE (“Sequel,” Prop, ix., p. 7).
Erect CD X to AB. CD is the required radical axis.
Dem.—Draw DG a tangent from any point D. Join DB.
Draw CE a tangent, and. join AE. Now AC2 — CB2 = AE2,
.·. AC2 - AE2 = CB2; that is, CE2 - CB2, .·. CE2 + CD2 = CB2
+ CD2; hut CE2 + CD2 = GD2 (“ Sequel,” Prop, xxi., p. 42),.
and CB2 + CD2 = DB2; .·. DG2 = DB2. Hence CD is the radical
axis (xyii., Ex. 6).
Sol.—Let C be the centre of the Θ, and A, B the points..
Join AB, and bisect it in D. Erect DE X to AB. Join AC,
and find the radical axis EE (Lemma) of the Θ and the point Ar
and let it cut DE in E. E is the centre of the required Θ.
Bern.—From E draw the tangent EG to the Θ. Join
EA, EB. EG = EA (Lemma), and EA = EB; .·. EA, EB, EG
are equal; and the Θ, with E as centre and EA as radius, will
pass through B, and cut the given 0 orthogonally in G
(“Sequel,” Book III., Prop. xxii.).
(2) Lemma.—To find the radical axis of two circles. Let
A, B he the centres. Join AB, and divide in E, so that AE2
- EB2 is equal to the difference of the squares of the radii.
Erect EG X to AB. From C and E draw tangents CD, EH,
CG, EJ to A and B. Join AH, BJ. Now AE* - EB2 = AH*book in.]
EXERCISES ON EUCLID.
129
- BJ2, EH2 = EJ2, .·. CE2 + EH2 = CE2 + EJ2; hence
(“Sequel,” Book III., Prop, xxi.) CD2 = CG2. Hence EC is
the radical axis of the circles.
Sol.—Let A, B be the centres, and P the point. Join AB,
and find the radical axis CE (Lemma). Join BP, and find the
radical axis CF of the point P, and the Θ B. From C, where
CE, CF intersect, draw tangents CD, CG to A and B. Join CP.
C is the centre of the required circle.
Dem.—Since CE is the radical axis of the Θβ A, B, CG = CD
(Lemma); and because CF is the radical axis of the Θ B and the
point P, CG = CP; .·. CG, CD, CP are equal, and therefore
the Θ, whose centre is C, and radius CP, will cut the Θ* A and
B orthogonally (“ Sequel,” Book III., Prop. xxi.).
(3) Let A, B, C be the G\ Find DE, the radical axis of A
and B, and DF the radical axis of A and C. From D, where
DE, DF intersect, draw tangents to A, B, C. Now these tan-
gents are equal; and the Θ, with D as centre, and one of them
as distance, will pass through the ends of the other two, and
will cut the Θ® A, B, 0 orthogonally (“Sequel,” Book III.,
Pfcop. xxi.).
x130	EXERCISES ON EUCLID.	[book iik
43.	Dem.—Join BD, AE, and let them intersect in 0. Join
CO, and produce it to meet AB in H.
Now (xxxi.) each of the L · ADB, AEB is right, .·. BD, AE
are Xs to AC, BC; hence (I., Ex. 16, Miscellaneous) CH is
X to AB. Now (xxu., Ex. 1) AHEC is a cyclic quadrilateral;
(xxxvi.) BC.BE = AB.BH. And since BHDC is a cyclic
quadrilateral, AC . AD = AB. AH. Adding, we get AC. AD
+ BC. BE = AB (AH + BH) = AB2.
44.	Dem.—Join AE, BF, CG, DH. Now(xxn.) the L%AEF,
ABF are together equal to two right L ·, and similarly the L *
AEH, ADH are together equal to two right L ■; hence the sum
of those L * is four right Ls, and the sum of the L · AEF, AEH,
FEHJis four right L ·; .*. the L FEH = ABF + ADH. In like
manner the L FGH = FBC + HDC; .·. the L * FEH and FGH
= ABC and ADC, and are equal to two right Le. Hence
(xxn., Ex. 1) EFGH is a cyclic quadrilateral.
45.	Dem.—Describe a Θ about ABC. Take any point M inHOOK in.]
EXERCISES ON EUCLID.
131
the circumference. Join MA, MB, MC. It is required to prove
that MA = MB + MC. Produce BM to D, so that MD = MC.
Join CD. Now (xxn.) the L* BAC and BMC are together
A
-equal to two right Z.8, and BMC, DMC together equal to two right
£·; .·. DMC = BAC, and is an L of an equilateral Δ ; and
because MC = MD, MCD is an equilateral Δ.
Again, because BMCA is a cyclic quadrilateral, the L MBC
= MAC, and ABC = AMC; but ABC = MDC, since each is an
J_ of an equilateral Δ ; .*. AMC = MDC; hence (I. xxvi.) the
Δ· AMC, BDC are equal; AM = BD; that is, AM = MB
4- MC.
46. (1) Let ABC be a Δ, the sum of whose sides AB, AC is
given, and the L BAC, both in magnitude and position. About
the Δ ABC describe a O. It is required to prove that the locus
of its centre F is a right line.
Dem.—Bisect the are BC in D. Join AD. Let fall a _L DE
D
on AB. From F let fall a J. FG- on AD. Now AE = h (AB
K 2132
EXERCISES ON EUCLID.
[boos, ieu
+ AC) (xxx., Ex. 4); hence AE is a given line; .*. E is a
given point. And since DE is X to AE, at a given point, DE is
given in position; and because the L BAD = J BAC, BAD is a
given L; .*. AD is given in position, and DE is given in posi-
tion ; .*. D is a given point, and the point A is given; hence AD
is a given line, and (in.) AD is bisected in G; .·. G is a given
point, and FG is JL to a line given in position; hence FG is
given in position. Hence the locus of F is the line FG.
(2) Bisect the L BAC by AD. Erect DE X to BC. DE is
J£
the diameter. Join EA, and from E, F let fall Xs EG, FH on
AB and AE.
Now the line AG is given, for it is equal to J (AB — AC);
.·. EG, which is X to it, is given in position, and EA is given
in position, since it is X to AD ; E is a given point, and EA
is bisected in H (in.); .*. FH is given in position. Hence the
locus of F is the line FH.
47. (1) Let 0 be the centre of the given G, and A, B the
points. It is required to describe a Θ which shall pass through
A, B, and bisect the circumference of the given Θ.
Sol.—Bisect AB in C. Join CO, and divide it in D, so that
CD2 — OD2 = R2 — BC2 (R being the radius of the given Θ).
Erect DE, CE, X8 to OC, AB, and join AE, BE, OE. E is the
centre of the required Θ.
Dem.—The Δ* ACE, BCE are equal (I. iv.); .·. AE = BE ;
hence the Θ, with E as centre, and AE as radius, will pass
through B. Let it cut the given Θ in G, H. Join OG, OHy
EG, EH. Nov CD2 - OD* = R2- BC2; .·. CE2 - OE* = R2BOOK III.]
EXERCISES ON EUCLID.
133
- BC2 ; .·. BC2 + CE2 = R2 + OE2 ; that is, BE2 = R2 + OE2;
„·. GE2 = R2 + OE2; but OG = R; .·. GE2 = OG* + OE2; hence
tthe L EOG is right. Similarly EOH is a right angle; OG
and ΟΞ are in the same straight line; hence GH is the diameter
of the given Θ. Hence the circumference of the given Θ is
bisected by the Θ ABH in the points G, H.
(2) Let A he the given point, and 0, O' the centres of the
given Θ·. It is required to describe a Θ passing through A
which shall bisect the circumferences of the Θ* whose centres
are 0, O'.
Sol.—Join AO, and divide it in B, so that AB2 — BO2 = R-
(R being the radius of the Θ whose centre is 0). Join AO', and
«divide it in C, so that AC* - CO'2 = R'2. Erect BD, CD JL* to
AO, AO'. Join AD. With D as centre, and AD as radius,134	EXERCISES ON EUCLID. [book m-
describe a Θ EAG, cutting the given Θβ in the points E, F r
G, H. This is the Θ required.
Dem.—Join OE, OF, O'G, O'H, OD, FD. Now AB2 - OB2
-	OF2 (const.), .·. AD2 - OD2 = OF*, .·. AD2 = OD2 + OF2;
that is, FD2 = OD2 -f OF2, .·. the L DOF is right Similarly,,
the L DOE is right, OE and OF are in the same straight line.
Hence EF is the diameter of one of the given Θ8. In like man-
ner GH is the diameter of the other given Θ. Hence the cir-
cumferences of the given Θ* are bisected by the Θ EAG.
48.	Let a Θ, whose centre is D, bisect the circumferences of
two given Θ8 in the points E, F ; G, H. It is required to find
the locus of D. (See last diagram.)
Sol.—Join EF, GH. Now since the circumferences are bisected
in E, F; G, H, the centres, must be in the lines EF, GH. Bisect
these lines in O, O'. Join 00', DO, DO'. From D let fall
a A DJ on 00'. DJ is the locus of D.
Dem.—Join DF, DH. Now (iii.) the L · DOF, DO'H are
right, .*. DF2 = DO2 + OF2, and DH2 = DO'2 + O'H2; but DF2
= DH2, .·. DO2 + OF2 = DO'2 + O'H2, .·. DO2 - DO'2 = O'H2
—	OF2; but O'H2 - OF2 is given, since O'H and OF are the
radii of two given Θ8, DO2 — DO'2 is given, .·. OJ2 — O'J2
is given; .*. J is a given point; .*. the line DJ is given in posi-
tion. Hence the locus of D is the line D J.
49.	Let 0, O', O" be the centres of the given Θ8; and R, R'r
R" their radii. It is required to describe a Θ which shall bisect
the circumferences of the given circles.BOOK. HI.]
EXERCISES ON EUCLID.
135
Sol.—Join 00', and divide it in G, so that OG2 — O'G2 = It'2
- R2. Join 0Ό'', and divide it in H, so that 0"H - O'H2
= R'2 — R"2; and at G, H erect GJ, HJ _L* to 00', O’O". The
point J, where these perpendiculars intersect, is the centre of the
required circle.
Dem.—Join OJ, O'J, 0"J. Through 0, O', 0" draw AB,
CD, EP at right angles to OJ, O'J, 0"J, and join JA, JB, JC,
JD JE, JF. Now OA2 = OB2, .·. OA2 + OJ2 = OB2 + OJ2;
.·. AJ2 = BJ2; .·. AJ = BJ. In like manner CJ = DJ, and EJ = FJ.
Again, OG2 - O'G2 = R'2 - R2; .·. 0G2 + R2 = 0'G2 + R'2, OG2
+ JG2 + R2 = O'G2 + JG2 + R'2; that is, OJ2 + R2 = O'J2 + R'*;
.·. AJ2 = DJ2 ; .·. AJ = DJ. Similarly, BJ = EJ, and CJ = FJ.
Hence those six lines are equal; and the Θ, with J as centre,
and AJ as radius, will pass through the points B, C, D, E, F,
and will bisect the circumferences of the given circles in those
points.
50. Dem.—Join BC, CD, DB. Now, since ABE is a right-
angled Δ, and BC is ± to AE, we have AE. AC = AB*
(1. Χιλή., Ex. 1). In like manner AF.AD = AB2; there-
fore AE.AC = AF.AD. Hence the points C, E, F, D are
ooncyclic.
51. (1) Dem.—Let J, K be the centres of the Θ·. Join JK,
and produce it. JK produced must pass through G (xi.) Join
KF. If GF does not pass through A, let it pass through
H. Now (xviii.) the L CFK is right, and the L CDB is right,
.·. FK and AB are parallel, .·. the L GFK = GHB; but GFKEXERCISES ON EUCLID.
136
[book hi.
= FGK (I. y.) ;	the L JHG = JGH; hence JG = JH; hut JG
e JA; .·. JH = JA, which is absurd. Hence GF produced must
pass through A.
(2) Complete the Θ ACD, and produce CD to meet the cir-
cumference again in M. Now (hi.) DC = DM, .·. the arc AC
= AM; hence (Ex. 26) AF.AG = AC2, and (xxxvi.) AF. AG
= AE2, .·. AC2 = AE2; .·. AC = AE.
52. Let ACB be an obtuse-angled triangle. It is required to
draw from C a line CE, so that CE2 = AE. EB.
Sol.—Describe a Θ about ACB. Let D be its centre. Join
CD.	On CD as diameter describe a Θ, cutting AB in E. Join
CE.	CE is the required line.
Dem.—Produce CE to meet the circumference again in F, and
join DE.
Now the L CED is right (xxxi.), .·. FED is right; hence (m.)
CF is bisected in E; .·. FE. EC = EC2; but (xxxv.) FE. EC
= AE.EB, .·. AE.EB = CE2.
53. Dem.—From E let fall 18 EF, EG on AB, CD, and join
AE, CE. Now AF = BF (hi.) ; .·. AB2 = 4 AF2. Similarly,
CD2 = 4 CG2; .·. AB2 + CD2 = 4 AF2 + 4 CG2. Again (I. xlvii.),BOOK III.]
EXERCISES OH EUCLID.
137
OE2 = OG2 + EG2 = EF2 + EG2 ; .·. 4 OE2 = 4 EF2 + 4 EG2.
Adding, we get AB2 + CD2 + 4 OE2 = 4 AF2 + 4 EF2 + 4 CG*
^ 4EG2; but 4AF2 + 4EF2 = 4 AE- = 4 R2, and 4CG2 + 4EG*
= 4 CE2 = 4 R2. Hence AB2 + CD2 + 4 OE2 = 8 R2.
54. (1) Let ABC be the triangle. From A, B, C let fall A “AD,
BE, CF on the sides, and intersecting in 0. It is required to
prove that AB2 + BC2 + CA2 is equal to 2 AO . AD + 2 BO . BE
+ 2 CO . CF.
Dem.—AC2 = AO2 + OC2 + 2 AO . OD (II. xn.), BC2
= CO2 + OB2 + 2 CO . OF, and AB2 = AO2 + OB2 + 2B0 . OE.
Adding, we get AB2 + BC2 + CA2 = (2 AO2 + 2 AO . OD) + (2 OB2
+ 2 OB . OE) + (2 CO2 + 2 CO . OF) = 2 AO (AO + OD)
+ 2BO (BO + OE) + 2 CO (CO + OF) = 2 AO . AD + 2B0.BE
+ 2 CO . CF.
(2) Describe a Θ about ABC, and from its centre Q let fall As
QG, QH, QJ on the sides, and join AQ. Now (iti.) AJ = BJ:
.·. AB2 = 4 AJ2 = 4 AQ2 - 4 QJ2; but AQ = R, and 2 QJ = OC138
EXEBCISES ON EUCLID.
[book hi.
(“ Sequel,” Book I., Prop, xii., Cor. 3); .·. AB2 = 4K*- OC*.
Similarly, BC2 = 4B2 - OA2, and CA2 = 4 K2 - OB2. Hence
AB2 + BC2 + CA2 = 12 K2 - (OA2 + OB2 + OC2).
55. Bern.—Join the centres O, O'. Produce DC, and let it
meet the circles again in the points G, H. Join OG, ΟΉ.
Now the L DCO' = OCG (I. xv.) ; but OCG= OGC ; .*. DCO'
= OGC, and O'DC = ODG (hyp.); .·. the Δ· ODG, O'DC are
equiangular; hence (xxxv., Cor. 3) OG . CD = O'C . DG. Again,
the L· O'HD, O'HC are equal to two right Z.*; and the L5
OCD, O'CD are equal to two right L *; and O'CD = O'HC;
the L O’HD = OCD, and (hyp.) O'DH = ODC; .*. the
Δ* O'HD, OCD are equiangular; hence (xxxv., Cor. 3) O'H.CD
= DH . OC. Multiplying these results, we get CD2 = DH . DG.
Now DG . DC s* DE2 (xxxvi.), and DH . DC = DF2;
.·. DG. DH . DC2 = DE2. DF2; .·. DC4 = DE2 . DF2 ; DC2
= DE. DF.KOOK IV.]
EXERCISES ON EUCLID.
13»
BOOK IV.
PROPOSITION IV.
1.	Bern.—CF = CD, OC common, and the base OF = OD ;
Renee (I. vm.) theZ. OCF = OCD. (Fig., Prop. iv.).
2.	Dem.—BD = BE, CD = CF, AE = AF(III. xvii.) ; .·. CB
+ AE = ^ (AB + BC + CA) = j j that is, λ + AE — s \ .*. AE
= (* — a). In like manner BD = (s — b), and CF = ($ — <;).
(Fig., Prop, iv.)
3.	Bern.—From O' let fall X8 O'F, O'G, O'H on the sides
AB, BC, CA -of the Δ ABC. Now, because the L 0'CG = 0'CH,
and the L O'GC = O'HC, and the side O'C common, .*. (I. xxvi.)
O'G = O'H. Similarly, O'G = O'F ; .*. O'F, O'G, O'H are equal,
and the 0 with O' as centre, and O'F as radius, will pass through
G and H.
4.	Let D, E be the points in which AC, CB produced touch
the Θ whose centre is O'". It is required to prove that BE
= (*- <*)·
Bern.—Let J be the point of contact of AB and O'". Now
it may be proved, as in Ex. 2, that CB + BJ = s; that is, CB
+ BE = s; but CB = a; hence BE = (* — »).
5.	(1) It is required to prove that the points 0, O'", A, B are
concyclic.
Bern.—Let E be the point in which BC produced touches 0'"·
Now since the L * ABC, ABE are bisected, the L OBO'" is equal
to half the sum of the L8 ABC, ABE, and is therefore a right
angle. Similarly, OAO'" is a right angle; the L8 OAO"',
OBG'" are together equal to two right angles. Hence (III. xxii.)
the points 0, 0", A, B are concyclic.
(2) It can be shown as in (1) that the Xs O'AO", O'BO" are
right Z.8. Hence (III. xxii., Cor. 1) the points O', B, A, 0" are
concyclic»
6.	It is required to prove that 0 is the orthocentre of the
Δ O'O'O ".HO
EXERCISES ON EUCLID.
[book IV.
Dem.—Because the L 0"B0"' is right, 0"B is the perpen-
dicular from 0" on O'OO". Similarly, 0Ά, 0"'C are the per-
pendiculars from O', 0"' on 0"0'", O'O". Hence the point O
is the orthocentre of the Δ 0'0"0'".
7.	See Book I., Miscellaneous Ex. 36.
8.	Dem.—It can be shown, as in Ex. 5, that the four points
O, A, O'", B are concyclic; hence (III. xxi.) the L A0"'0=AB0;
but ABO = CBO ; .-. CBO = AO'"C, and the L ACO'" = BCO,
since ACB is bisected; hence (I. xxxii., Cor 2) the Δ· BOC,
ACO'" are equiangular; .*. (III. xxxv., Cor. 3) CO.CO'"
= BC . AC = ab. In like manner AO . AO' = be, and BO . BO"
= ca.
10.	Dem.—From 0' let fall J_8 r' on AB and AC produced,
and on BC join O'A, O'B, O'C. Now br — 2 Δ ACO', cr' = 2 ABO';
.·. r'(b -f c) = twice the quadrilateral ACO'B, and ar = 2 BO'C ;
.·. r'(b + e — a) = 2 ABC ; but (a 4- b + e) = 2s; (i + e — a)
- 2 (s - a); .·. 2r' (s — a) = 2 ABC. Hence r' («-«) = area of
the Δ ABC.
11.	From 0, 0' let fall _L8 OK, O'H on AC. It is required to
prove that OK . O'H = ($ — b)(s — c); that is, rrl = (s — b)(s — <?).
Dem.—The line AH = s (Ex. 4), and CH, CK are (* — b) and
(s — e) (Exs. 4 and 2). Now the L OCO' is right (Ex. 5), .·. the
L8 OCK, O'CH are together equal to a right L ; and since the
L O'HC is right, the L8 HO'C, HCO' make together a right Z ;
.·. the L HO'C = OCK, and the L O'HC = OKC, each being right;
.·. the Δ8 O'HC, OKC are equiangular. Hence (III. xxxv.,
Cor. 3) OK. O'H = (s — b) (s — e) ; that is, rr' = (s - b) (# — e).
12.	Dem.—Area of Δ ABC = rs (Ex. 9), and r'{s — a) = area
of ABC (Ex. 10); .*. r.r'.s.s — a = square of area of ABC;
but rr' — s — b.s — c (Ex. 11). Hence square of area of ABC
=	e — b . 8 — e.
13.	Dem.—Let the area of ABC be denoted by Δ. Now rs —A
(Ex. 9), and r^.s — a = Δ (Ex. 10). Similarly r". s— b = Δ, and
r'".s — c = a; hence (**.*Λ*·".*·"')(«.8 —	b.s — e) = Δ2; but
(s.s—a.s — b.s — c) = A2 (Ex. 12). Therefore r. r'.r". r " = Δ2.
14.	Dem.—From O'" let fall X8 0"'D, 0"'D' on CB, CA. Now
in the Δ 0"'D'C the L 0"'D'C is right, and the ZD'CO"' is half
a right L\ ·*. the L CO"'D' is half right;	(I. vi.) D'O"^ D'C;
-but DO'" = r" and D'C = s (Ex. 4); .·. r'" = s. Similarly it can
be shown, if we let fall ±8 OE, OE' from 0 on CB, CA, that E'CBOOK IV.]
EXERCISES ON EUCLID.
141
= E'O; but ΕΌ = r, and E'C =(« — <?) (Ex. 2); r - (* — e).
In. like manner r* = (s — b), and r" = (s — a).
15. (1) Let AB be the base, CDE the vertical Z, and R the
radius of the inscribed circle. It is required to construct the
triangle.
Sol.—Produce CD to F, and bisect the Z EDF by DG. On
AB describe a segment of a Θ containing an Z = CDG. Erect
BJ X to AB and = R. Through J draw JH || to AB, and cut·
ting the Θ in H. Join AH, BH, and let fall a _L HI on AB„
At the points A, B, in the lines AH, BH, make the Ls HAK,
HBL respectively equal to the Zs HAB, HBA, and produce AKr
BL to meet in N. ANB is the required triangle.
Dem.—From H let fall ±8 HM, HL on AN, BN. Now in
the Δ* HIB, HLB we have the Z8 HIB, HLB = HLB, HBL,
and the side HB common; .·. (I. xxvi.) HI = HL. Similarly
HI = HM; hence the Θ with H as centre, and HI as radius,
will pass through L and M, and its radius = R, for HI
= BJ = R.
Again, the Z8 of the Δ HAB are equal to two right Zs, and
the L8 CDG, FDG are equal to two right Z8; but the Z AHB
= CDG, .*. the Z FDG = HAB + HBA; and because the Xs of
the Δ ANB are two right Z8, .*· the L8 of ANB are equal to the
Z* CDG, FDG; but the Z8 NAB + NBA = 2 (HAB + HBA)
= 2 FDG = FDE. Hence the remaining L ANB = CDE.
(2) Let AB be the base, CDE the vertical L, and R the radius
of the escribed Θ which touches the base and one of the sidts
produced.
Sol.—Bisect the Z CDE by DF, and on AB describe a seg-
ment containing an Z = CDF. Erect BJ X to AB and = R..142
EXERCISES ON EUCLID.
[book nr.
Through J draw JH || to AB, and from. H, where it meets the Θ,
let fall a A HK on AB produced. With H as centre, and HK
as radius, describe a Θ. From A, B draw tangents to this Θ,
meeting in N. ANB is the required triangle.
Dem.—Join AH, BH. Now HK = JB = R, and because H
is the centre of the escribed Θ of the Δ ANB, AH, BH are the
bisectors of the internal L NAB and the external L NBK (I.
xxxii., Ex. 12), the L AHB = JANB; hut AHB = ^CDE.
Hence ANB = CDE.
(3) Let AB be the base, CDE the vertical L, and R the radius
of the escribed Θ which touches the base externally and the sides
produced.
Sol.—Produce CD to F, and bisect the L EDF by DG. On
AB describe a segment of a Θ containing an Z = EDG. ErectBOOK IV.]
EXERCISES ON EUCLID.
14?
BJ JL to AB, and make it equal to R. Through J draw JK || to
AB, and cutting the Θ in K. From K let fall a X KL on AB.
With K as centre, and KL as radius, describe a Θ. Through
A, B draw tangents IN, MN to this Θ, meeting in N. ANB is
the triangle required.
Dem.—Join KA, KB. Since K is the centre of the escribed
Θ of the Δ ABN, the L AKB = £ (NAB + ABN) (I. xxxii.,
Ex. 12); but AKB = JFDE (const.); .·. NAB + ABN = FDE;
hence the L ANB = CDE, and LK, the radius of the escribed
Q, = BJ = R.
PROPOSITION V.
2.	Dem.—Because each of the L * APB, AQB is right, AQPB
is a cyclic quadrilateral, and AP, BQ are chords in the O; hence
(III. xxxv.) OA. OP = OB . OQ. Similarly OB. OQ, = OC. OR.
(Diagram 2, Ex. 1.)
3.	Dem.—The L AOF = DOC (I. xv.), and AFO = CDO,
each being right; FAO = OCD; but OCD = GAF (III. xxi.);
FAO s= GAF, and AFO = AFG, each being right, and AF
common. Hence (I. xxvi.) OF = GF. (Diagram, Ex. 1.)
5.	Dem.—In the Δ ΟΌ'Ό'" the lines O'A, 0"B, 0"'C are
X· from O', 0", O'" on 0"0'", 0"O', O'O" (iv., Ex. 6), and the
points A, B, C are the feet of these Xs; hence (Ex. 4), the Θ
about ABC is the nine-points Θ of the Δ ΟΌ'Ό'". In like
manner it is the nine-points Θ of the Δ* 00Ό", 00"0'",
ΟΟΌ'". (Diagram, Ex. 3, Prop, iv.)
6.	Dem.—Because the lines IF, IH, IK are equal (Ex. 4),
and the L KFH is right, HK is the diameter of the Θ about the
Δ KFH; .*. IK, IH are in one straight line; and since KH is
H to CP, and CK to PH, PCKH is a parallelogram; .*. CK = PH;
but CO = 2 CK, CO = 2 PH. (Diagram, Ex. 4.)
7.	Dem.—IF = JPG. This is proved in Ex. 4.144
EXERCISES ON EUCLID.
[book IV.
PROPOSITION X.
1.	Dem.—The L ACD = CBD + CDB (I. xxxn.); but CBD
= 2 CAD (x.), and CDB = CAD. Hence the L ACD = 3 CAD.
2.	Dem.—The L8 of the Δ ABD are equal to two right L 8; but
each of the L8 ABD, ADB is equal to 2 BAD; hence the L BAD
is £ of two right L8; that is, of four right Z.8; .·. the arc
BD is of the whole circumference. Hence the line BD is a
side of a regular decagon.
3.	Dem.—Let A be the centre. Join AB, AD, AE, AF, and
join BF, cutting AD in G. Now since BD is a side of a regular
inscribed decagon, ABD is an isosceles Δ, having each of its base
L8 double of the vertical L (Ex. 2), the L BAD is I of two-
right Z8, .*. the L BAF is £ of two right L8; hence the L AFB
is I of two right L % .·. the L AGF is f of two right L8; .·. AF
= GF, that is, BF — BG = R. Now the L DBG is £ of two right
L8, and BDG is f, .·. BGD is £, .·. BG = BD. Hence BF — BD
= R.
4.	Dem.—Because ACDE is a cyclic quadrilateral, the L8
ACD, AED are together equal to two right Z* (III. xxii.) ; and
the L8 ACD, BCD are together = to two right L8, .·. the L AED
= BCD; that is, AED = CBD; but AED = ADE, and CBD
= ADB, ADE = ADB, and AD common. Hence (I. xxvi.)
DE = DB.
Again, the L ACE = ADE (III. xxi.), and the L CDA = CEA mT
but (x.) CDA = CAD = DAE; .·. CEA = DAE, and the side
AE « AD. Hence (I. xxyi.) the Δ· ACE, ADE are con-
gruent.
δ. Dem.—Let 0 be the centre of the Θ ACD. Join OA, OC.
Now (Ex. 4) AEC is an isosceles Δ, having each base L double
of the vertical L; and since the L8 of the Δ AEC are together
equal to two right Z·, the L AEC is £ of two right L8; hence
(III. xx.) the L AOC is f of two right L8; that is, £ of four
right L 8. Hence AC is the side of a regular pentagon.book iv.] EXERCISES ON EUCLID.
145
PROPOSITION XL
1.	Let ABODE be a regular pentagon inscribed in a Θ, and
letits diagonals CE, AD intersect in A'; BD, CE in B'; CA, BD
in O'; AC, BE in D'; and BE, AD in E'. It is required to prove
that A'B'C'D'E' is a regular pentagon.
. Dem.—Because the arc AE « BC (xi.), the L ECA * BAG,
.·. CE is || to AB; hence (I. xxix.) the L8 EB'B, B'BA, are
together equal to two right La; for the same reason the L8 CA'A,
A'AB are equal to two right L8; but the L DBA = DAB; hence
the L A'B'B * B'A'A. In like manner the L8 at C', D', E' are
equal. Hence the figure A'B'C'D'E' is equiangular.
Again, because the arc BC = DE, the L BDC = DCE ; .·. the
aide B'C = B'D, and (I. xv.) the L CB'C' = A'B'D; and the
L B'C'C = B'A'D, because they are the supplements of the
equal L8 B'C'D', B'A'E'; hence the side C'B' = A'B'. Similarly,
the other sides of A'B'C'D'E' are equal. Hence it is a regular
pentagon.
2.	Produce AE, CD to meet in A'; ED, BC in B'; DC, ABin
C'; CB, EA in D'; BA, DE in E'. Join A'B', B'C', &c. It is
required to prove that A'B'C'D'E' is a regular pentagon.
Dem.—In the Δ8 ABD', CBC', the Z ABD' = CBC', and
the L D'AB = BCC', being the supplements of equal Z8, and the
side AB = CB; hence (I. xxvi.) BD' = BC', and the L AD'B
= BC'C. Similarly, AD' = AE'; EE' = EA'; DA' = DB'; and
CB' = CC'. Again, because the Z ABC = EAB, the Z D'BA
= D'AB; .*. D'A = D'B. Now in the Δ8 D'AE', D'BC#, we
have the sides AD', AE' = BD', BC', and the contained Z8 equal;
hence the base D'E' = DO'. In like manner the other sides are
equal. Hence the figure is equilateral. Again, we proved the
Z BD'C' = BC'D', and the Z AD'B =BC'C ; and the Z AD'E is
= CC'B', since the Δ8 AD'E', CC'B' are equal in every respect.
Hence the Z E'D'C' = D'C'B'. In like manner the other Z8 are
equal. Hence the pentagon A' B'C'D'E' ib regular.
3.	Let AD, BE, two consecutive diagonals of a regular pen-
tagon, intersect in E. It is required to prove that EB. EE'
= E'B2.
Dem.—Join CE, and describe a Θ about the Δ AE'B.
Now because DE = BC, the L DCE = BEC ; .*. DC is || to BE.
L146
EXERCISES ON EUCLID.
[book it.
Similarly, BC is || to AD ; hence (I. xxxiv.) DC = BE'; hut DC
=|AB (hyp.): .*. AB = BE'. Again, because AE = DE, the
L ABE = EAD, and hence (III. χχχιι.) AE is a tangent to the
Θ ABE'; .·. (III. xxxvi.) EB.EE' = AE* = AB* = ΈΤΒΚ
Hence BE is cut in extreme and mean ratio in E\
4.	Let AB be a side of a regular pentagon. It is required
to construct it.
Sol.—Erect BC _L to AB, and make it equal to J AB. Join AC,
and produce it to D, so that CD = CB. On AB describe an
isosceles Δ ABE, haying each of its equal sides equal to AD.
About the Δ ABE describe a Θ. Bisect the Z8 BAE, ABE by
the lines AF, BG, meeting the circumference in F and G. Join
AG, GE, EF, BF. ABFEG is the required pentagon.
Bern.—From AC cut off CH = CB or CD. Now DA. AH
+ CH2 = AC2 (II. vi.); but CH2 = BC2, and AC2 = AB2 -f BC2;
.*. DA. AH = AD2 = DH2; .·. AD is divided in extreme and
mean ratio in H. Therefore, since AE = AD, if we divide AE
in extreme and mean ratio, the greater segment would be equal
to AB, and hence (x.) AEB is an isosceles Δ, having each base
L double the vertical L; but the base Z.8 are bisected by the
lines AF, BG; .*. the L s EAF, FAB, ABG, GBE, AEB are equal;
.·. the chords EF, BF, AG, EG, AB are equal. Hence ABFEG
is a regular pentagon.
5.	Let ABC be a right L. It is required to divide it into five
equal parts.
Sol.—Draw BD, making the L ABD equal to the vertical L
of an isosceles Δ having each of its base Z.8 double the vertical
L. Bisect the L ABD by BE; each of the Z8 ABE, DBE is £ of
a right L. Draw BF, BG, making the Zs DBF, FBG each
equal to EBD. Then the L ABC is divided into five equal
parts by the lines BE, BD, BF, BG.
PROPOSITION XV.
1. (1) Let ABCDEFbe the hexagon. Join, AC, AE, CE. It is
required to prove that the area of the hexagon is double the area
of the Δ ACE.
Dem.—Let the diagonals of the hexagon intersect in 0. NowBOOK IV.]
EXERCISES ON EUCLID.
147
the Δ8 OCD, OED are equilateral, and hence OCDE is a lozenge,
*nd CE is its diagonal; .*. OCDE = 2 0CE. Similarly OACB
2 OAC, and OAFE = 2 OAC. Hence ABCDEF = 2 ACE.
(2) See Book I., Prop, i., Ex. 4.
2.	Let AB he the diameter, and 0 the centre. Produce AB to
•C, so that BC = BO. From C draw tangents CD, CE to the 0,
.and join DE. It is required to prove that the Δ CDE is equi-
lateral.
Bern.—Join OD, OE, BD, BE. Now (III. xvm.) the L CDO
is right; .·. (Book I., Prop, xii., Ex. 2) the lines BD, BO, BC
are equal; hut OB = FO; .*. the Δ ODB is equilateral; and
because each of the Z8 CDO, CEO is right, CDOE is a cyclic
quadrilateral, .·. the Z8 DOE, DCE are together equal to two
tight Z8; hut each of the L8 DOB, BOE is an L of an equi-
lateral Δ, .·. DCE is an Z of an equilateral Δ ; and because
CD = CE, the Δ CDE is equilateral.
3.	(1) Let ABCDEF be the hexagon, and GHJ the equilateral
Δ. It is required to prove that the area of the Δ is double the
.area of the hexagon.
Bern.—Let the diagonals of the hexagon intersect in 0. Join
AG, CH, EJ. Now, because AB = AF, AG common, and the
base GB = GF, .·. (I. vm.) the L BAG = FAG, and the L OAB
= OAF; .*. the Z8 FAG, OAF are together equal to two right Z6;
hence (I. xrv.) OA and AG are in the same straight line.
Again (III. xvm.), the Z OFG is right, .·. the Z8 FOG, FGO
make one right L ; but the Z AFO = FOA; .·. the L AFG143
EXERCISES ON EUCLID.
[book m
= AGF; .·. AF = AG; but AO = AF; .*. AO = AG; Hence
(I. xxxvii.) the Δ AFO = AFG; .*. the Δ OFG = 2 OFA.
Similarly, OBG = 2 OBA; . ·. OFGB = 2 OFAB. In like maimer
OBHD = 2 OBCD, and OFJD = 2 OFED. Hence the Δ GHJ
= 2 ABCDEF.
(2) Let A'B'C'D'E'F' be the circumscribed hexagon. It is re·
quired to prove that the area of ABCDEF is three-fourths the
area of A'B'C'D'E'F'.
Dem.—Because each of the L8 F'AO, F'FO is right (III.
xvm.), the L * AF'F, AOF are together equal to two right L *r
and the L8 AF'F, AF'G are together equal to two right Ls 'r
hence the L AF'G = AOF; .·. AF'G is an L of an equilateral A.
In like manner AA'G is an L of an equilateral Δ ; GF'A' is
an equilateral Δ ; and because GA is _L, it bisects the base;.
.·. AF' = AA'; .·. A'F' or GF' = 2 AF' = 2 FF'; hence tho
Δ F'GA = 2 FF'A ; .·. FGA = 3 FF'A ; hence (1) AOF
= 3 FF'A; .·. AOF = f OFF'A. In like manner AOR
= | OAA'B, &c. Hence ABCDEF = f A'B'C'D'E'F'.
Exercises on Book IV.
1. (1) Let ABODE be a regular polygon circumscribing a Θ.
It is required to prove that the corresponding inscribed polygon is-
regular.
Dexn.—Let O be the centre. Join OF, OG, OH,. 01, OJ.
Now (III. xvm.) the L* OHD, 01D are right, .·. the-book nr.]
EXERCISES ON EUCLID.
149
Z8 IDH, IOH are together equal to two right Z8. In like
manner the Z * GCH, GOH are together equal to two right Z 8;
but IDH = GCH (hyp.); .·. the Z IOH = GOH. In the same
way it can be shown that all the Z8 at 0 are equal. Hence the
arcs are all equal, and therefore the five chords FG, GH, HI, IJ,
JF are all equal.
(2) Proved as in Book IV., Prop. xii.
2.	Let the Δ ABC be isosceles. It is required to prove 'that
the Δ DEF is isosceles.
Bern.—Let 0 be the centre. Join OD, OE, OF. Now
the Z8 ODB, OEB are right (III. xviii.), .*. the Z8 EBD,
EOD are together equal to two right Z8. Similarly the Z*
FCD, FOD are together equal to two right Z8; but the Z EBD
= FCD (hyp.); .·. the Z EOD = FOD; .·. the arc ED = FD;
-·. the chord ED = FD. And hence the Δ DEF is isosceles.
3.	Let the Z BAC = EDF. It is required to prove that both
Δ8 are equilateral.
Dem.—Because the Δ8 are isosceles, and the Z BAC = EDF,
their remaining Z8 are equal; .·. the Z ABC = EFD ; but EFD
= EDB (III. xxxii.); EBD = EDB, and EDB = BED;
EBD is an Z of an equilateral Δ. Similarly FCD is an
Z of an equilateral Δ. Hence ABC and DEF are equilateral
triangles.
4.	Let ACB be an Z of an equilateral Δ. It is required to
•divide it into five equal parts.
Sol.—Describe a Θ about the Δ ABC, and in it inscribe a
tegular polygon of fifteen sides (xvi.); then five of those sidesΙδΟ	EXERCISES ON EUCLID.	[book it.
will be in the arc AB. Let D, E, F, G be the points of division.
Join CD, CE, CF, CG. Now since the arcs AD, DE, EF, FG*.
GB are equal, the Z.8 ACD, DCE, ECF, FCG, GCB are equal.
5. Let ABC be a sector of a given circle. It is required to·
inscribe a circle in it.
Sol.—Bisect the L BAC by AD, meeting the arc BC in D.
Through D draw EF a tangent to the sector. Produce AB, AC
A
to meet EF. Bisect the L AEF by EG, meeting AD in G. G is-
the centre of the required circle.
Dem.—From G let fall ±* GH, GJ on AE, AF. Now (III.
xviii.) the L EDG is right, and the L EHG is right (const.)»
and the L DEG = HEG, and EG common; .·. (I. xxvi.) GD
= GH. Similarly GH = GJ. Hence the Θ, with G as centre
and GD as radius, will pass through H and J.BOOK IV.]
EXERCISES ON EUCLID.
161
6.	Dexn.—Describe a Θ about ABC, and through A draw AF
touching this Θ. Now (III. xxxii.) the L FAC = ABC; but
ABC = ADE (I. xxix.); .·. FAC * ADE; .·. the Θ about ADE
will touch AF in A. Hence the circles touch each other in A.
7.	Bern.—Let EF be the tangent at E to the circle about ABE.
Now the L FEA = EBA (III. xxxii.); butEBA = DCA (III.
xxi.). Hence the L FEA = DCA, and therefore the lines EF,
CD are parallel.
D
8.	Let ABCD be a given square. It is required to describe a
regular octagon in it.
Sol.—Draw the diagonals AC, BD, intersecting in 0. Cut off
AE, AF« AO ; BI, BJ = BO; CG, CH = CO; DK, DL = DO.
Join EG, JL, HF, XI. EGJLHFKI is the octagon required.
Dem.—Join OE, OG, 01. Now, because AE = AO, and the
L EAO is half a right Z.,	each of the L8 AEO, AOE is three-EXERCISES ON EUCLID,
[book it.
n%
fourths of a right L, and the L AOB is right; EOB is one-
fourth of a right Z. Similarly, each of the L8 GOB, AOJ is one-
fourth of a right L ; hence EOI is half a right L, and we have
seen that AEO is three-fourths of a right L; EIO is three-
fourths of a right L; .*. 01 = OE. And because the L EOB
= GOB, and EBO = GBO, and the side BO common, OG = OE
= 01. Now OG = 01, and OE common, and the L GOE = IOE ;
the bases EG, El are equal. In like manner all the sides are
equal. Again, because BE = BG, the L BEG = BGE ; each
is half a right L; each of the L8 GEI, EGJ is three halves of
a right L . In like manner all the L3 are equal. Hence the
octagon is regular.
9. Let AB, AC be two given lines, and BC a line of given
length sliding between them. From B, C X8 BD, CE are let fall
on AC, AB, intersecting in F. It is required to find the locus
of F.
Sol.—At B, C erect Xs BG, CG to AB, AC. Join FG, cut-
ting BC in H. Join AF, AG, AH. Now, because CE and GB
are perpendicular to AB, and CG, BD to AC, CGBF is a paral-
lelogram; hence (I. xxxiv., Ex. 1) BH = CH, and FH = GH.
Again, since BC is a line of given length sliding between two
fixed lines, AB, AC, and BG, CG are perpendiculars at its ex-
C
tremities; (III. xxvm., Ex. 2) the locus of G is a Θ, having
A as centre, and AG as radius; hence AG is a given line, and
(I. xlvii.) AC2 + CG2 = AG2, and AB2 + BG2 = AG2; .·. AC2
+ CG2 + AB2 + BG2 is given; but (II. x., Ex. 2) BG2 + CG2
= 2 CH2 + 2 HG2, and AB2 + AC2 = 2 CH2 + 2 AH2; .*. 4 CH*
+ 2 AH2 + 2 GH2 is given; but 4 CH2 = CB2; .'. 4 CH2 is
given, and 2 AH2 + 2 GH2 is given; .·. AF2 4- AG2 is given;hook ιν.]
EXEBCISES ON EUCLID.
15$
but AG2 is given, .*. AF is given; hence AF is a line of given
length; and since A is a fixed point, the locus of F is a Θ having
A as centre, and AF as radius.
10. Let ABC be the triangle. About ABC describe a Θ. Let
DF be a X at the middle point of AB. Produce DF to meet the
circumference in E. Join AD, BD, CD, CE. It is required to
prove that CE is the internal, and CD the external bisector of the
L ACB.
Bern.—Produce BC to H, and join AE, BE. Now, be-
cause AF = BF and FE common, and the L AFE = BFE,
the base AE is equal to BE; .*. the arc AE = BE; hence the
L ACE = BCE. Therefore CE is the internal bisector of the
L ACB. Again (I. iv.), AD = BD, and the L FAD = FBD;
and becauee ABCD is a cyclic quadrilateral, the L ■ BAD and
BCD are together equal to two right Ls, and the L8 BCD, DCH
are together equal to two right L*; the L BAD = DCH,
and (III. xxi.) the L ACD = ABD, and ABD « BAD; hence
ACD = DCH. Therefore CD is the external bisector of the
L ACB.
11. Sol.—Draw any tangent AB to the Θ. At the point B
make the L ABC = X. From the centre 0 draw OD X to BC;
and with O as centre, and OD as radius, describe a Θ. Through
P draw a* tangent to this Θ, cutting BCE in E and F. PEF is
the line required.
Dem.—Take any point G in BCE, and join BG, CG, EG, FG.164
EXERCISES ON EUCLID.
[book it*
Now (III. kit.) EF = BC, .·. the L EGF = BGC; but (III*
xxxn.) BGC = ABC = X. Hence the L EGF = X.
F
12. Let IJK be the given Θ ; A, B the given points, and’CD
thejgivsn line. It is required to inscribe a Δ in IJK, having
two of its sides passing through A, B, and the third parallel
to CD.
Sol.—Join BA, and produce it to meet CD. Produce EB to
F, and make AB. BF = EB. BG. Through F draw FHI, cutting
the O in Η, I, and subtending an L = CDF (Ex. 11). Join IB,
and produce it to meet the Θ in J. Join JA, and produce it to
meet the Θ in K. Join IK. IJK is the required triangle.
Bern. — Join HK. Now, because AB.BF = EB.BG,.
.·. AB.BF = JB.BI; hence the points A, J, F, I are concyclic,
-·. the L AJI = AFI; but AJI = KHI (III. xxi.); .·. AFIEXERCISES ON EUCLID
155
κοοκ nr.]
= ΚΗΪ. Hence AF and EH are parallel; and since the L HKI
~ CDF, IK is | to CD.
13. Let A, B, C, D be four points, no three of which are col-
linear. It is required to describe a Θ which shall be equidistant
from them.
8ol.—Describe a Θ passing through A, B, C. Let 0 be its
centre. Join OD, cutting the Θ in E. Bisect ED in F. With
O as centre, and OF as radius, describe a Θ GHI. This is the
Θ required.
Dem.—Join OA, OB, OC, and produce them to meet the
Θ GHI. Because OF = 01, and OE « OC; .·. EF = Cl;
but EF = DF, .·. Cl as DF. In like manner BH and AG are
equal to DF. Hence the Θ through G, Η, I, F is equally dis-
tant from the points A, B, C, D.
14. Let ABC be a given Θ. It is required to inscribe a Δ ia
A
it, whose sides shall pass through three given points D, E, F-150
EXERCISES ON EUCLID.
[book IV.
Analysis.—Let ABC be the Δ. Join EF. Through B draw
BH parallel to EF, and meeting the Θ in H. Join AH, and
produce it to meet EF produced. Join GD. Through H draw
HJ parallel to GD, and meeting the Θ in J. Join JB, and pro-
duce GD^to meet JB in I. Now because BH is || to EG, the
L CEG’= CBH; but CBH = CAH (III. xxi.); .·. CEG = CAH ;
hence ECGA is a cyclic quadrilateral; EF . FG = AF . FC;
but AF . FC is given, since F is a given point; EF . FG is
given; but EF is given, FG is given; .*. G is a given point.
Again, the L ABJ = AHJ (III. xxi.); but AHJ = AGI
(I. xxix.), .*. ABJ = AGI; hence BIAG is a cyclic quadrilateral;
.·. GD . DI = AD . DB; but AD . DB is given, .*. GD. DI is
given, .·. DI is given; hence I is a given point. Now the
L JHB = IGE (I. xxix., Ex. 8); but the L IGE is given, since
the lines EG, IG are given in position; .*. the L JHB is given.
Hence the question reduces to Ex. 11.
15. (1) Let R be the radius of the inscribed circle; X the
vertical L, and P the perpendicular from the vertical L on the
base. It is required to construct the triangle.
Sol.—With any point A as centre, and a radius equal to R,
describe a Θ. Draw BC a diameter, and at the point A in AB
make the L BAD = X. Through C, D draw EF, EG tangents
to the 0. With E as centre, and a radius equal to P, describe
a Θ. Draw FG a common tangent, touching the Θ8 in K, L.
EFG is the required triangle.
Dem.—Join EL. Now the L ELF is right; EL is the _L from
the vertical L on the base, and it is equal to P (const.); and AD,
the radius of the inscribed 0, is equal to R. Again, each of the
Z* ACE, ADE is right (III. xviii.) ; .*. the Z.8 CAD, CED areBOOK IV.]
EXERCISES ON EUCLID.
137
together equal to two right /.*; and the L · CAD, BAD are
together equal to two right L *; CED = BAD = X.
(2) Let X be the sum of the sides, Y the base, and R the
radius of the inscribed circle.
Sol.—Take a line AB, and from it cut off BC = \ (X + Y) and
CD = Y. Erect DE _L to AB, and make it equal to R. With E.
B
as centre, and ED as radius, describe a Θ. DrawBG, touching
this Θ at F. Join BE, and produce it. Erect CH _L to ABr
and meeting BE produced in H. From H draw HG 1 to BG.
With H as centre, and HC as radius, describe a Θ. Draw IJ a
common tangent, touching the Θ8 in K and L. BIJ is the
required triangle.
Dem.—ED, the radius of the inscribed Θ, is equal to R; and
(iv., Ex. 2) IJ + BD = i (IB + BJ + IJ), and (iv., Ex. 4) BC
= £(IB + BJ + IJ); hence IJ = CD = Y. Again, BC = J(IB
+ BJ 4- IJ), and BC = £ (X + Y) (const.); hence (IB -f B J + IJ>
= (X + Y); (IB + BJ) = X.
(2') Let X be the base, Y the difference of the sides, and R
the radius of the inscribed circle.
Sol.—With any point 0 as centre, and a radius equal to R,
describe a Θ. In the circumference take a point D. Through I>
draw a tangent BC. From DB cut off DG = Y. Bisect DG in
H, and make BH, CH each equal to J X. Through B, C draw
AB, AC tangents to the circle. ABC is the triangle required..168	EXERCISES ON EUCLID; [book it.
Dem.—BC = BH 4- CH = X; and AB = AF + FB, and AC
= AE + EC. Hence AB - AC = FB - EC = BD - CD = BD
— BGr = DG = Y.
A.
R
(3) (Diagram, Prop, iv., Ex. 3).—Let O', O", O'" be the centres
«of the escribed circles. Join them, and let fall XB O'A, 0"B,
0"'C, on O'O'", 0'"0', O'O". Join AB, BC, CA. ABC is the
triangle required.
Dem.—Produce AB, AC to F and H. Let 0 be the point
where the X* intersect. Now because each of the L ■ 0"C0'",
O AO" is right, AOCO" is a cyclic quadrilateral; .*. the L ACO"
= AOO". Similarly the L BCO' = BOO'; but AOO" = BOO',
and ACO" = O'CH, .·. BCO' = O'CH; hence CO' is the external
bisector of the L ACB. Similarly, BO' is the external bisector
of the L ABC. Hence O' is the centre of the escribed Θ touch-
ing BC externally and the other sides produced. In like maimer
O", O'" are the centres of the other escribed circles.
16. (1) Dem.—From D let fall X* DH, DJ on AC and OB
produced. Join DA, DB, HF, FJ; the points H, F, J are col-
linear (III. xxii., Ex. 12). Join DC, CE, and through F draw FG
parallel to DC. Now because the L ACB is bisected by CD,
HC = \ (AC + CB) (III. xxx., Ex. 4); and since the L DHC is
right, DC . CO = HC2 (I. xlvii., Ex. 1); that is, DC. FG = HC2.
Again (III. xxxi.), the L DCE is right; .*. EGF is right, and
CLD is right; .*. EGF = CLD, and (I. xxix.) the L EFG
= LDC; .·. the Δ8 DCL, EFG are equiangular; hence (III.
xxxv., Cor. 3) DC. FG = DL, FE ; DL. FE = HC*.BOOK IV.]
EXERCISES ON EUCLID.
15»
(2) From C let fall a X CK on AB. Now (III. Ex. 17)
FM. FK ie equal to the square of half the difference of AC and
CB; that is, equal to AH2.
Again, the L ELC = DFM, each being right; and because
DCE is right, the L * CED, CDE are together equal to a right L ;
and the L ■ LEC, LCE are equal to a right L ; .·. LCE = CDE;
hence the A* DFM, CLE are equiangular, .·. (III. xxxv., Cor. 3)
DF. LE = LC . FM = FK. FM = AH2.
17. Let the regular polygon of n sides be a pentagon ABODE,
P a point within it, and pi, p2> &c., the A* from P on the sides.
Let 0 ho the centre of the inscribed Θ, and R its radius. It is
required to prove that (pi + p% + pz + pi + p5) = 5 R.
Dem— Join AP, BP, &c., and let the sides be denoted by i.
How spi = twice the A APB ; tp%=twice the A BPC, &c.; hence160	EXERCISES ON EUCLID.	[book nr.
* (Pi +JP2 + pz + j>4 + p&) ~ twice the area of the pentagon.
Again, eR = twice the Δ AOB, &c.,	5 sR = twice .the area
of the pentagon; .·. s (pi + p^ + p3 + p* + p5) = 5 «R. Hence
(Pi + Pi + pz + P4, -f ρδ) = 5 R. Similarly for a regular polygon
of any number of sides.
18. Let A, B, C, D, E be the angular points of a regular
polygon of five sides. About ABODE describe a Θ ; and through
A, B, C, D, E draw tangents to this 0. It is required to prove
that the sum of the Xs from A, B, C, D, E on any line is equal
to five times the X from O, the centre of the Θ, on the same
line.
(1)	Bern.—Let the line be a tangent at any point P in the
circumference. From P, C let fall Xs PF, CG on the tangents
through C and P. Produce CF to meet PG in H. Now in the
Δ8 CGH, PFH, the L CGH = PFH, and PHC is common, and
the side CH = PH; hence (I. xxvi.) CG = PF. Similarly,
theX8 from A, B, D, E on the tangent at P are equal to the
X8 from P on the tangents through those points; but (Ex. 17)
the sum of the X* from P on the sides of ABCDE is equal to
5R; hence the sum of theXs from A, B, C, D, E on PH
= δ R; that is, equal to five times the X from 0 on PH; and
similarly for a regular polygon of any number of sides.
(2)	Let the line not touch the circle.
Bern.—From A, B, C, D, E let fall X8 AA", BB", CC", DD",
EE" on the line, and from 0 let fall 00". At O', where 00" cut»
the Θ, draw a line parallel to C"E", and let the X8 from A, B,
C, D, E cut this line in A', B', C', D', E\ Now (1) AA' 4- BB'161
book iv.] EXERCISES ON EUCLID.
+ CC' + DD' -f- EE' b 600', end A'A + B'B + C C" + DO"
+ ΕΈ" = δ 0Ό". Hence, by addition, we get AA" + BB" + CC"
+ DD" + EE" = 5 00".
19. (1) Let ABC be a Δ inscribed in a O. Bisect the
base AB in F, and erect a 1, meeting the circle in D, E;
then (III. m.) DE is the diameter. Join^CD, CE. CD and CE
are the external and internal bisectors of the L ACB (Ex. 10).
Produce AB to Ο, H. Bisect the L8 CBG, CAH by BO', AO",
meeting CD produced in O', 0". Produce O'B, 0"A to meet in
M162
EXERCISES ON EUCLID.
[book it.
O'".	O', 0”, O'" are the centres of the escribed Θ8 (iv., Ex. 3).
Produce CE. CE produced will pass through O'" (I. xxvi., Ex. 8).
From O'" let fall a X 0"'J on AB. Join AO', meeting CE in 0.
From 0 draw OK parallel to 0"J, and from O'" and E draw
0"'K and EL parallel to AB. From O', 0" let fall 1* O'G (r'),
0"H (r") on GH. Join BE.
Now, since AO', BO', CO' meet in O', and that BO', CO' are
two external bisectors, hence (I. xxvi., Ex. 8) AO' is the internal
bisector of the L BAC. Similarly, BO" is the internal bisector
of the L ABC.
Again, AG, BH are each equal to s (tv., Ex. 4); AH = BG;
.·. HF = GF; hence HG is bisected in F; .·. (I. xl., Ex. 8)
O'G + 0"H = 2 DF ; that is, r + r" = 2DF. And because the
L ECB = ACE, .·. (III. xxi.) ECB = ABE, and CBO * ABO:
hence (I. xxxn.) EOB = EBO; .·. EB = EO; but the L OBO"'
is right, .·. the L* BOO", B0"0 are together equal to a right Z ;
but EOB = EBO; .·. EO'"B = EBO'"; .·. EB = EO"'; hence
00'" is bisected in E, and EL is parallel to 0"'K; .·. (I. xl.,
Ex. 3) OK is bisected in L, and divided unequally in M; hence
KM - OM = 2 LM; that is, r"' — r = 2 EF; and we have proved
r'+r"=2DF;.*.r' + r" + r'" - r = 2 DE = 4 R. Hence / + r"
+ r'"= 4R + r.
(2)	It has been shown that r'" — r = 2 EF; but EF is = μ; hence
r'" — r = 2μ. Similarly r — r = 2μ\ and r" — r = 2μ"; hence
r' + r" + r'" — Zr = 2 (μ + μ' + μ"), that is, 4 R + r - Zr
= 2 (μ + μ* + /a"). And therefore (μ + μ* + μ") = 2 R — r.
(3)	Dem.—μ + 8 = R, μ + 6' ■= R, and μ” + 3" = R; hence we
have μ + μ + /a" + δ + S' + S" = 3 R; that is, 2R-r + S + S/
+ 3" = 3 R. And hence « + 3' + 3" *= R + r.BOOK. IT.]
EXERCISES ON EUCLID.
163
20. Dem.—Let G be the second point of intersection. Join
GB, GC, GD. Now (III. xxii.) the sum of the L* DGC, DEC
it two right L *; but DEC - EAF + AFE, and AFE = BGI)
(III. xxi.); BGC + BAC is equal to two right /·; hence
BACG is a cyclic quadrilateral; the circle through B, A, C
will pass through G. And the locus of G is a circle.
21. Let ABCD be the quadrilateral, E, F the middle points of
the diagonals, and O the centre of the inscribed Θ. It is required
to prove that the points E, O, F are collinear.
Dem,—Join EP>, ED, and join 0 to the points of contact
G, Η, I, J.164
EXERCISES ON EUCLID.
[book IV*.
Now (I. xxxvra.) the Δ ABE = CBE, and the Δ ADE = CDE ;
AEB + CDE = J ABCD; hence the sum of the areas of AEB
and CDE is given, and their bases AB, CD are given; .·. (i.,
Ex. 29) the locus of E is a straight line, andF is a point onthelocus
since it can be shown in the same manner that AFB + CFD =
J ABCD. Again, the Δ OAG = OAH, and OIB = OBH; .·. the
area of OAB is half the area of the figure GABIO. Similarly, OCD·
= f GOICD ; hence OAB + OCD = \ ABCD, and .*. (i., Ex. 29)
0 is a point on the locus; that is, the points E, 0, F are on the
same straight line.
22. (1) Let AB he a given line; C, D two given points. It is re·
quired to find a point P on AB, so that CP + DP = R (a given line).
Sol.—With C as centre, and a radius equal to R, describe a]0,
and describe a second Θ DEF, having its centre on AB, passing
through D, and touching the first Θ internally in E (III. xxxvn.r
Ex. 3). Let P be its centre. P is the requiredjpoint.
Dem.—Join CP, and produce it; then (III. xi.) CP pro-
duced passes through E. Join PD. Now PE = PD; .·. CP*
+ PD = CE = R.
(2) It is required to find a point P, so that CP - DP = R.
Sol.—With C as centre, and a radius equal to R,Jdescribe a QrSOOK ΙΥ.]
EXERCISES ON EUCLID.
165
and describe a second Θ DEF, having its centre on AB, passing
through D, and touching the first Θ externally in E. Let P he
its centre. P is the required point.
Dem.—Join CP, DP. Now CP = CE + EP; .-.CP - EP
= CE = R ; that is, CP - DP = R.
23.	Let AB, AD, CB, CF he the four lines. About the
Δ· AFE, CDE describe Θ8; let them intersect in P. P is
the point required.
Dem.-—From P let fall Is PG, PH, PI, PJ on the sides of
the Δ® AFE, CDE.
Now (III. xxii., Ex. 12) the feet of the J_8 on the sides of the
Δ AFE are collinear. Similarly the feet of the X* on the sides
<of the Δ CDE are collinear. Hence the feet of the Xs PG, PH,
PI, PJ are collinear; and these are the Xs on the four given
lines AB, AD, CB, CF.
24.	See “ Sequel to Euclid,” Book III., Prop, xiv*
25.	See “ Sequel to Euclid,” Book III., Prop, xiv., Cor.
26.	(1) See “ Sequel to Euclid,” Book III., Prop. v.
(2) Dem.—Let AB be the diameter of the semicircle ACB.
Produce BA to D, and let a Θ whose centre is 0 touch ACB in
■C, and’JBD in D. From 0 let fall a X OD on BD. Join 00'.
OO’ passes through C (III. xii.). Produce 00' to meet ACB in
E, and from 0, D draw OF, DG tangents to ACB. Now
EO. OC = OF2 (III. xxxvi.) = OD2 + DG2 (“ Sequel,” Book III.,166
EXERCISES ON EUCLID.
[book it,
Prop, xxi.), and OC2 = OD2. Subtracting, we get (EO - OC) OC *
that is, EC . OC « DO2; that is, 2 Rr = DG2 = DA. DB.1
27. Lemma.—If a Δ ABC have a Θ inscribed in it, and another
circumscribed to it, the rectangle contained by the diameter of
the circumscribed Θ and the radius of the inscribed Θ is equal to-
the rectangle contained by the segments of any chord of the circuni-
circle passing through the centre of the inscribed circle.
Dem.—Let 0 be the centre of the inscribed Θ. Join AO, and
produce it to m eet the circumscribed Θ in D. From D let fall a
X DF on BC, and produce it to meet the circumference in E..
Join EC, OG, OC, BD, CD. Now the arc BD = CD; the-
chord BD = CD; hence BF = CF; .·. DE is the diameter of
the circumscribed Θ; .*. the L DCE is right, and (III. xvm.)
the L OGA is right, and (III. xxi.) the L DEC = OAG ;.HOOK IV.]
EXEECISES ON EUCLID.
187
hence the Δ* DEC, 0A6 are equiangular; (III. xxxv.,
Cor. 3) ED. OG « OA. DC ; hut DC = DO (Dem., Ex. 19 (1)).
Hence ED . OG = OA. OD.
Let ABC he the Δ, 0, O' the centres of the inscribed and cir-
cumecribed Θ", and p, p'the radii of two O*, touching each other
at 0, and touching the circumscribed Θ. It is required to prove
112
that - +	r being the radius of the inscribed circle.
Ψ 9 r
Dem.—Through 0 draw a common tangent to those Θ·, meet-
ing the circumscribed Θ in D, E. Join the centres of the Θ*
whose radii are p, p\ and produce to meet the circumferences
in I, J. Through O’ draw FH || to IJ, and cutting DE in G.
Now FG.2p = EO.OD (“Sequel,” Book III., Prop, vi.) ;
™ EO.OD 0. M n	EO.OD M EO.OD
··■PG = -57-· Slmilarly·HG=—qr; ·’■FH = ~w
+	· Again, 2 Rr = EO. OD (Lemma); .·. 2 R =	;
2 p	r
EO.OD EO.OD EO.OD ,	.	1	11
. ·.-----t- ---:— = -------: therefore -— + r—. = -;
2p	2p'	r	2 p 2/ r
11 2
p p r
28. Lemma.—Let AB be the diameter of a Θ, AD a tangent·168
EXERCISES ON EUCLID*
[book it.
From C, any point in the circumference, a JL CD is let fall on
D, and AC joined. It is required to prove that AB.CD = AC2.
Dem.—Through C draw CE || to AD. Join BC. Now
B
(I. ΧΙΛΊΙ., Ex. 1) AB.AE = AC2; but AE = CD ; .·. AB.CD
= AC2.
Dem.—Let the polygon be a regular pentagon ABCDE. Take
any point F in the circumference. At F draw a tangent to the 0.
Join F to the angular points of the pentagon, and let the joining
C
lines be denoted by c\, <?2> &c. From the angular points let fall
-1“ pit piy &c., on the tangent, and let the radius be denoted
by R.
Now" {Lemma) 2 Rpi = cr, and 2 R= <?«r, &c. ; .·. 2 R (pi
· . . .Pb) = (ci2+ C2* .... c52); but^+^2 ....ρ6) = δΚ
(Ex. 18) ; .·. 10 R2 = (cf2 + <"22 . - . . c£). And similarly for a
figure of any number of sides.BOOK IT.]
EXERCISES ON EUCLID.
169
29.	This is a special case of the next exercise.
30.	If any point G in the circumference of any concentric O
he joined to the angular points of an inscribed regular polygon,
the sum of the squares of the joining lines is equal to n times the
square of the radius of the concentric Θ, together with n times
the square of the radius of the circumscribed Θ ; that is, pi2 + p22
+ . . . . p52 = δ R2 + 5 r2.
Dem.—Let 0 be the common centre. Through 0 draw the
G
diameter. From A let fall a JL pi on IJ, and draw AK parallel
toIJ.
Now AG2 = OG2 + OA2 — 2 OG. OK (II. xiii.) ; that is, pr
= R2 -t- r2 - 2 Rpi. Similarly, p22 = R2 + r2 + 2 ILpt, &c., the sign
of 2 R?2 being positive, since the _L is let fall from above the
line. Adding, we get, since the terms by which 2 R is multiplied
cancel each other, pi2 + p22 + . . . . ps2 = δ (R2 + r2).
31.	Let ABC be an equilateral Δ inscribed in a Θ. From P,
any point in the circumference of a concentric Θ, _L8 pi, Pi, pz,
are let fall on the sides of ABC. It is required to prove that
Pi2 + P22 + Ps2 = three times the square of the radius of the in-
scribed Θ, together with three half times the square of the radius
of the concentric circle.
Dem.—From 0, the centre, let fall _L8 on the sides of ABC.
Through P draw PD || to AC, meeting the ± from 0 on AC
in D; draw PE |j to BC, meeting the _L from 0 on BC in E.
Produce the _L from 0 on AB to F, and draw PF || to AB.
Join OP. Now, since the L8 ODP, OEP, OFP are right, the170
EXERCISES ON EUCLID.
[book it.
Θ on OP as diameter will pass through D, E, F; and be-
cause PD is || to AC, and PE || to BC, (I. xxix., IJx. 8) the
L DPE = ACB = an L of an equilateral Δ ; DE is J of the
circumference of DEF. In like manner, EF, DF are each J of the
circumference of DEF ; D, E, F are the angular points of an
equilateral Δinscribed in DEF, and .·. (Ex. 28) OD2 + OE4 + OF*
(OP\ 2	3 OP2
—J = —-—. Again, pi — (r — OD), r being the radiua
of the inscribed Θ ; .·. pi2 = r2 — 2 r . OD + OD2 = (r2 4- OD2)
- 2 r (r — pi)9 and piz = (r2 + OE2) — 2 r (r — p%)y and p$*
3 OP2
=* (r* + OF2) - 2 r (r - p$) ; .·. pi2 + pz2 + pz2 = 3 r2 +
- 2r {3r - (pi +p* + pz)} ; but (pi +pz + p*) = 3r (Ex. 17).
3 OP2
Hence j?i2 + p*? + pi2 = 3 r2 -i-—-. And in general, in the case
o a figure of n sides, the sum of the squares of the i.8 will equal
β «ΟΡ2
"•+-r·
32 & 33. These are special cases of Ex. 31.KOOK IT.]
EXERCISES ON EUCLID.
m
35. Let A, B, C, D be the four concyclic points. From O, the
centre of the Θ, let fall 1· Oo, 00, Ογ, Οδ on the sides of
ABCD ; then (III. ni.) the sides of the quadrilateral are bisected
in a, 0, y, 5. From a, y let fall X* oE, 7F on AB, CD, and let
them intersect in O'. Join 00', and produce it to meet AD in G.
It is required to prove that 00 is X to AD.
Dem.—Join ay, 05, 00'. Now, because o, 0, y, 5 are the
angular points of a parallelogram (I. XL., Ex. 6), and that aOyO*
is a parallelogram; .*. (I. xxxiv., Ex. 1) the lines ay, 05, 00^
bisect each other. Let P be their common point. Now, in the-
Δ* 0P5, OT0 we have the sides OP and P5 equal to O'P and
P0, and the contained L% equal; the L 05P = O'0P; .·. 00
is |] to 05; .·. 0G is X to AD. Similarly, if we join 50' and
produce it, 5H will be X to BC.
36. Let MN be an arc of a Θ whose centre is 0. Let AB be
O
Die side of a regular pentagon, and A'B' the side of a regular172
EXERCISES ON EUCLID.
[book rr.
hexagon circumscribed about it. It is required to prove that the
perimeter of the pentagon is greater than that of the hexagon.
Dem.—Let AB touch MN in P. Join 0A, OB, OA', OB', OP.
Now (xii.) the Δ AOP = BOP; .·. the L AOP = JAOB; but
AOB =
ΑΌΡ =
4 rt. L*
5	;
2 rt.L*
AOP =
2 rt. L8
5
In like maimer the L
·. the L AOA' =
2 rt.L*
30	*
Let C be the point where 0 A' cuts the Θ. Then if we divide
the arc CP into five equal parts in the points D, E, F, G, join
•OD, &c.s and produce to meet AB in the points D', E', F', G',
the L 8 A'OD', DOE', &c., will be each of two right L *.
Again, the line OA is greater than OD' (I. xix., Ex. 4). Cut off
OH = OD'. Join A'H. Then (I. iv.) A'D' = ΑΉ, and the
L OD'A' = OHA'; .*. the L OD'E' = AHA'; but OD'E' is
greater than OAD' (I. xvi.); .·. AHA' is greater than A'AH;
and hence AA' is greater than A'H; that is, than A'D'. Simi-
larly, A'D' is greater than D'E'; D'E' greater than E'F', &c.;
hence 5 AA' is greater than A'P. To each add 5 A'P, and we
have 5AP greater than 6A'P; .*. δΑΒ is greater than βΑ'/Β';
but δ AB is the perimeter of the pentagon, and 6 A'B' that of the
hexagon. Hence the perimeter of the pentagon is greater than
that of the hexagon; and in general the greater the number of
sides, the less the perimeter.
37.	By the last exercise the area of a pentagon is less than the
area of a square; but the area of a square is equal to the square of
the diameter. Hence the area of a pentagon is less than the square
of the diameter.
38.	Dem.—Join the four concyclic points A, B, C, D. Bisect
the joining lines in E, F, G, H. Join BD, and bisect it in I.
Then (v., Ex. 4) the nine-points Θ of the Δ ABD will pass
through the points Η, E, I. Similarly, the nine-points Θ of the
Δ ABC will pass through E, F, and the middle point of AC.
Hence two of the nine-points Os will pass through E. In like
manner two of them will pass through each of the points F, G,
H. From E, F, G, H let fall ±8 EE', FF', GG', HH' on the
opposite sides; these JLS will co-intersect in a point 0 (Ex. 3δ).
Join IF, IG. Now, because each of the L8 AGO, AF'O is right,
.·. the L8 F'AG', FOG' are together equal to two right L8, and
the L8 BAD, BCD are equal to two right .·. the L FOG'
— BCD; that is, the L FOG = BCG; but (I. xxxrv.) BCGBOOK IV.]
EXERCISES ON EUCLID.
173:
= FIG; .·. FOG = FIG; and hence the Θ through the points F,
G, I, must pass through 0. In like manner each of the four nine»
points Θ8 must pass through 0. Now, since two of these Θ· pass
through E and 0, if we bisect EO, and erect XXi ± to it, their
centres must be in ΧΧχ. Similarly, the centres of each other pair
must he in the lines X1X2, X2X3, X3X1· Hence the points X,
Xi, X2, X3 must he the centres. And because each of the lines
XXi, CD is ± to EE', they are parallel to each other. Similarly,,
the remaining sides of XXiX2X3 are parallel to the remaining
sides of ABCD ; hence the L · X and X2 are equal to the L ·
and C ; hut A and C are together equal to two right L»,
and X» are equal to two right L *. Hence the points X, Xi, X2,.
Xs are concyclic.
39. Let AB, AC be two fixed lines, haying their included.
A
/ BAC equal to an L of an equilateral Δ ; and let BC be a'
174
EXERCISES ON ETJCLID.
[book ITl
third line forming a Δ with AB, AC. Bisect BC, AC, AB in
D, E, F. Join DE, EF, DF. The Θ through D, E* F is the
nine-points Θ of the Δ ABC (v., Ex. 4). It is required to prove
that the locus of its centre 0 is a right line.
Bern.—Join OA, OE, OF. Now DE, DF are respectively
| to AB, AC (I. XL., Ex. 2); .*. AEDF is a parallelogram:
.·. the L FDE = FAE; but FOE = 2 FDE (III. xx.); .·. FOE
= 2 FAE; hence FOE is twice an L of an equilateral Δ ;
.·. FOE + FAE are equal to two right Z*; hence FAOE is a
cyclic quadrilateral. Again, because OE = OF, the arc OE
*= OF, and (III. xxyi.) the L OAE = OAF ; .·. the L FAE
is bisected. Hence the line OA is given in position; and since 0
is a point on it, the locus of 0 is a right line.
41. Let AB, AC he two lines given in position, P a given
point, and let the line FG be equal to the given perimeter. It ie
required to draw a transversal through P, so that the Δ DAE
•hall have a perimeter equal to FG.
Sol.—Bisect FG in H. In AB take AB = GH, and erect BO
X to AB. Bisect the L BAC by AO, and let fall a JL OC on
AC. Then (I. xxvi.) the Δ* ABO, ACO are equal in every
respect; .*. OB = OC; hence the Θ, with 0 as centre and OB
as radius, will pass through C, and touch the lines AB, AC in
B, C. Through P draw DE, touching this Θ, and cutting AB,
AC in D, E. ADE is the Δ required. For (rv., Ex. 4) AB is»oox it·] EXERCISES ON EUCLID.
176
-equal to half the perimeter of ADE. Hence the perimeter if
•equal to 2AB, or FG.
42. (1) Let BAC he the vertical L, X its bisector, and FG the»
perimeter.
Sol.—Bisect the L BAC by AP, and make AP = X. Through
P draw DE, cutting off a A ADE whose perimeter is equal to
FG (Ex. 41).
(2)	Let BAC be the vertical L , FG the perimeter, and X the
perpendicular.
Sol.—Bisect FG in H, take AB = GH, erect BO X to AB,
bisect the L BAC by AO, and from 0 let fall OC JL on AC;
then the Θ, with 0 as centre, and OB as radius, will pass through
O, and will touch AB, AC, in B, C. With A as centre, and a
radius equal to X, describe a Θ, cutting AB, AC in Μ, K.
Draw a common tangent to the two Θ8, meeting AB, AC in
D, E. ADE is the required triangle.
Dem.- Join AP, P being the point where DE touches the
Θ MN. Now (III. xviii.) the L APE is right, AP is a X,
and it is equal to X; and, as in Ex. 41, the perimeter of the
A ADE = FG.
(3)	Let BAC be the vertical L , FG the perimeter, and R the
radius of the inscribed circle.
Sol.—Bisect BAC by AO. Draw AD X to AC, and make if
D
equal to R. Through D draw DE || to AO, and where it meets
AC draw EH | to AD. From H let fall HI X on AB. Take AB176
EXERCISES ON EUCLID.
[book nr.
= \ FG; erect BO _L to AB, and from O let fall a _L OC on AC.
Now, as in Ex. 41, HE = HI, and OB = OC; hence the Θ8 with
Η, O as centres, and HE, OC as radii, will pass through the
points I, B. Draw a common tangent, touching the Θ8 in K and
L, and cutting AB, AC in Μ, N. AMN is the required triangle.
For, as before, the perimeter of AMN = FG. And since
ADEH is a parallelogram, EH = AD = R.
43. (1) Let ABC he the given Θ, D, E the points. It is re-
quired to inscribe a Δ in ABC, so that two sides may pass
through D, E, and the third he a maximum.
Sol.—Describe a Θ passing through D, E, and touching ABC
in A (III. xxxvn., Ex. 1). Join AD, AE, and produce to meet
ABC in B, C. Join BC. ABC is the required triangle.
Dem.—Take any other point F in ABC. Join FD, FE, and
produce to meet ABC in GH. Join GH, JE, J being the point
where FG cuts the Θ ADE. Now the L DJE is greater than
DFE; .·. the L DAE is greater than DFE; .·. the arc BC is
greater than GH. Hence the chord BC is greater than GH.
(2) Let ADE he the given Θ ; B, C the points.
Sol.—Through B, C describe a Θ ABC, touching ADE in A.
Join AB, AC, cutting the Θ ADE in D, E. xJoin DE. ADE is
the required triangle.
Dem.—Take any point F in ADE. Join BF, CF, cutting the
G ADE in GH. Join GH. Produce CF to meet ABC in J.
Join BJ. - Now the L BFC is greater than BJC, that is, greaterEXERCISES ON EUCLID.
BOOK IV.]
177
than BAC ; .*. the arc GH is greater than DE. Hence the chord
GH is greater than DE.
44. Let Δ represent the area of the triangle.
Now r'= —(iv., Ex. 10), r"= —; .*. r'r" =
s — a	s — o
hut Δ2 = s.s — a .8 — b .8 — e (iv., Ex. 12) ;
A2
(«-«Η»-*)·
therefore rV
8.8 —a.s —b.8 —c
(»-a)(s-6)	-··*-<>·
r’W — 8.8— b. Hence r V' + r"r
but (a + 5 + c) = 2 s (iv., Ex. 2); .·.
= 8.8=8*.
Similarly, r,f r’" = s.s — a, and
+ r"’ / = 8 (3 s — (a 4- b + r)};
r'r" + rV" + r"Y = s {3 s - 2 * }
45. Let ABC be a Δ inscribed in a Θ. Draw the diameter
DE. Join AD. AD is the internal bisector of the vertical l.
H178
EXEBCISES ON EUCLID·
From A let fall a 1 AJ on BC. From B and C let fall Xs BG,
CF on AD, and let H be the point where DE bisects BC. It is
required to prove that the points F, H, G, J are coney cli6.
Dem.—Join FH, GJ, CD. Now, since each of the L8 BGA,
BJA is right, BGJA is a cyclic quadrilateral; .*. the L BAG
= BJG. And because DHFC is a cyclic quadrilateral, the L DCH
= DFH ; but (III. xxi.) DCH = BAD ; .*. DFH = BJG. Hence
the points F, H, G, J are coneyclic.
46. Let ABC be the Δ whose base AB and vertical L ACB
are given.
Describe a Θ about ACB. Let 0 be its centre. Draw DE,
the diameter, perpendicular to AB. Join CD, CE. CD, CE
are the internal and external bisectors of the L ACB (III. xxx.,
Ex. 2). Bisect the external L CAB by AO", meeting CE pro-
duced. Produce CD, 0"A to meet in O"'. Join Or"B. Produce
0"'B, 0"C to meet in O'. O'B is the external bisector of the
L CBA (I. xxvi., Ex. 8); O', 0", O'" are the centres of the
escribed Θ®. Join O'A, 0"B, intersecting CD in F. Join FO.
Draw EG || to CD, meeting FO produced in G. G is the centre
of the Θ passing through O', 0", O'". It is required to find its
locus.(BOOK IV.]
EXEBCISES ON EUCLID.
179
Dem.—Join BD. Now, because F is the orthocentre of the
Δ 0'0"0'" (IV. iv., Ex. 6), 0 the centre of its nine-points Θ
(IV. v., Ex. 5), and EG the _L from the middle point of ΟΌ",
OF = OG (IV. v., Ex. 4); and since the L GEO = FDO
(I. xxix.), and GOE = FOD ; .·. EG = DF ; hut DF = DB
(Dem. of Ex. 27), and DB is given, .·. EG is given, and the
point E is given. Hence the locus of G is a Θ, having E as
centre and EG as radius.
47. Let ABC be the Δ ; O', 0", O'" the centres of the
escribed circles.
Dem.—Describe a Θ about the Δ ΟΌ'Ό'". Let 0 he its
^centre. Join 0"0, and produce it to meet the circumference in
D, and cutting AC in E. We shall prove that O'O is _L to AC.
Join 0"B, and produce it to meet the circumference in F. Join
DF. Now the L 0"FD is right (III. xxxi.), and 0"B0"' is
right, since 0"B is _L to O'O'";	O'O'" and FD are parallel;
(III. xxvi., Cor. 2) the arc 0"'D = O'F; hence the L 0"'0"D
= OO"F, and the L 0"AE = 0"0'B (i., Ex. 36); .·. the
L 0"EA = 0"B0'; but 0"B0' is right, 0"EA is right;
hence 0"0 is _L to AC. Similarly, if we join O'O, 0"'0,
they will be _L to BC, AB. Hence the three _L8 are con-
current.
48. -Dem.—Join BE, BQ, BD. Now (III. xxxi.) the L AQB
is right, and EOB is right (hyp.); .·. OEQ.B is a cyclic quadri-
lateral ; the L OQB = OEB ; but OQB = PAB (III. xxi.);
sr2180
EXERCISES ON EUCLID. [book i
D
therefore OEB is equal to PAB. And hence the Θ through tho
points A, E, B will pass through D.
49. Let ABC be a Δ whose sides a, bt c are in arithmetical
progression ; a being the greatest, and e the least. It is required
to prove that 6 Rr as ac.
Dem.—From B let fall a _L p on b, and draw the diameter.
Now 2 Up = ac (III. xxxv., Ex. 1),	2 R.ftp « abc ; but bp is
equal to twice the area; that is, equal to 2Δ (suppose),	2R. 2 A.
= abc, .*. R =	; and since the sides are in A.P., a + c = 2 b ;
.·. a + b + e = 34; but (a + b + c) = 2s, therefore 2* = 3$.
A
Again (iv. Ex. 9), r = —; and multiplying this and the equation
R = — we get Rr =*	.*. Rr = ^ ; and hence 6 Rr = ac.
4 A	4 s	ο o
50. Let A'B'C' be the Θ ; and AB, BC, CA three lines in the
form of a Δ. It is required to inscribe in A'B'C' a Δ whose
«ides shall be parallel to the sides of ABC.
Sol.—Tate a point A’ in the circumference, and draw A'B' f| to-βοοκ IV.]
EXERCISES ON EUCLID.
181
AB, and A'C' || to AC. Join. B'C\ If B’C' is || to BC, the thing
required is done. If not, from the centre 0 let fall a A OD on
B'C'. With 0 as centre, and OD as radius, describe a Θ. Draw
EF, touching this Θ, and ]| to BC (III. xvi., Ex. 2). Join O to
G, the point of contact. Draw FH || to C'A', and join HE.
HFE is the Δ required.
Dem.—Because OG = OD, EF = B'C' (III. xiv.), .*. the art
EF = B'C'; hence the arc FC' = B'E; but FC = HA' (III.
xxvi., Cor. 2), .*. ΒΈ = HA', .*. HE is parallel to A'B'; that is
parallel to AB; and FH is parallel to A'C', that is to AC; and
EF is parallel to BC. Hence the sides of the Δ HFE are parallel
tto the sides of ACB.
ol. Dem.—Let 0 be the centre of the circumscribed Θ. Join
D182
EXEBCISES ON ETJCLID.
[book IV.-
DA", CE", OA, OB, OE, &c. Now the Δ A'OE" + DOC
— (A'OC + E"OD) = 4 A'OE' (Book I., Ex. 52); that is, AA"E"’
+ AOA" + ACE" + DOC - (BOC + ΑΌΒ + EOD + EOE")
= 4 A'OE'; hut evidently AOA" = A"OB, AOE" = EOE", and
DOC = EOD ; .·. AA"E" - BOC = 4 A'OE', and BOC = A'OE'
H- A'AE'. Adding, we get ΑΑ'Έ' - A'AE' -f 5 ΑΌΕ' = pentagon
A'B'C'D'E'.
52.	(1) Dem. — Let ABCDE be the equilateral inscribed
polygon.
Now, since the sides are equal, the arcs are equal; therefore
the whole arc EABC = DEAB; hence the L CDE = BCD.
Similarly, the L BCD = ABC, &c. Hence the polygon i&
regular.
(2) Dem. — Let ABCDE be the equilateral circumscribed
polygon; F, G, Η, I, J the points of contact, and 0 the centre..
Join OA, OB, OF, OG, OH.
Now ID = HD, .·. IE = HC, .·. JE = GC, .·. AJ = BG,
.·. AF = BF. Now since AF = BF, OF common, and the L
AFO = BFO, .·. the L OAF = OBF; .-.the L BAE = ABC.
Similarly all the L2 * * * * * 8 are equal. Hence the polygon is regular.
53.	(1) Let ABCDE he the equiangular circumscribed polygon;
F, G, Η, I, J the points of contact, and O the centre. Join OA,
OB, OG, OH.
Now since the L CBA = EAB, their halves are equal; that
is, the L OBF = OAF, and the L OFB = OF A, each being right,
and the side OF common, .·. (I. xxvi.) BF = AF ; that is,
AB = 2 AF. Similarly, AE = 2AJ; hut AF = AJ, .*. AB
= AE. In like manner all the sides are equal. Hence the-
polygon is regular.
(2) Dem.—Let ABCDE be the inscribed polygon, and 0 the
centre. Join OA, OB, OC, OD, OE. Now the L ABC = EAB
(hyp.); hut the L OBA = OAB, since OA = OB, therefore the
L OBC = OAE ; that is, OCB = OEA; but the L BCD = AED,
OCD = OED; that is, ODC = ODE. Now, in the Δβ ODC,
ODE, the L8 OCD, ODC are equal to the L8 OED, ODE, and
the side OD common; hence (I. xxvi.) DC = DE. Similarly all
the sides are equal. Hence the polygon is regular.book iv.]	EXERCISES ON EUCLID.	183
54.	The sum of the _L8 drawn to the sides of an equiangular
polygon X from any point P inside the figure is constant.
Bern.—Suppose a regular polygon Y of the same number as X
constructed so as to include X, and have its sides parallel to those
of X. Then, if the ±8 from P on the sides of X he produced to
meet the sides of Y, their sum is constant (Book IV., Ex. 17);
hut the excess of the latter sum over the former is constant.
Hence the former sum is constant.
55.	Express the sides of a Δ in terms of the radii of its escribed
circles.
If the radii be r\ r”, we have, denoting the area of th$
triangle by A (Book IV., Prop, rv., Ex. Id),
.·. r (r" + /") -	(a-b) + (* - a)(s - e) ’
but (Book IV., Prop, nr., Ex. 12) A* = s.s - a.s - b.s - c;
.·. r'(r" + /") = s.s — c + s.s — a = sa,
and (Book IV., Ex.) Vr' /' + /' /"+/'' / = s;
________r* (/' + /")
*’* ^VfV' + r'V"+/">·''
r" (/"+/)'
Similarly,	b - ^=======-,184
EXERCISES ON EUCLID.
[book t.
BOOK Y.
Miscellaneous Exercises·
1. (1) Let a be greater than b. It is required to prove that
*	* is greater than
b — x
Dem.—Subtract, and we get
ab — bx—ab + ax
b(b-x)
; that is
(a-b)*
b(b-x);
§-x
but since a is greater than b, ^ -- —* is positive. Hence
b{b -z)	b-x
.	a
is greater than
b
a	a + x
(2) To prove that — is greater than ,---
b	b + x
_ e ,,	.	,	« a + z ab + ax - ab - b*
Dem.—Subtract, and we get 7-------------=-------—------------
b + x	b {b 4- x)
«=	; but because a is greater than b. [**- is positive.
δ (b + x)	b{b + x)
a	a + x
Hence - is greater than ,----.
b	6	b + x
2.	The proof of this is similar to that of Ex. 1.
3.	Let a, bf c, d be the four magnitudes; then if a: b : : e : d>
a + b c + d
it is required to prove that
- b e-d'
Dem.—Because a : b : : c : d, we have a + b : b : : c + d : d
(xvm.); that is —Again, a-b:b::c-d:d (xvxi.);
b	d
that is,
a — b c — i
Dividing, we get
a + b c + d
b	d	a — b c — d’
4. Let a, b, c, d, and e,ft g, h, be the two sets of four magni-
tudes that are proportionals; that is, a: b:: e: d; and e:/::g: A.
It is required to prove that ae : bf: : eg : dh.ROOK V.]
EXERCISES ON EUCLID.
185
Dem.—Because a : b : : c : d9 we have 7 = -· Similarly,
0 a
J — Multiplying together, we get — = —j; that is, ac : bf
:: eg : dh.
* .	.	„	, f! b e d
5.	It is required to prove that - : y : : ~
a £	eg	a e c
Dem.—As in (4), we have = and - = ^--ry— -
g ch__c d a
h ~ dg g ' h * ‘ ’ *
g a e af «	b	and	e d ~
bed 	 a	b	e	d
■f- = -TT. Hence - :			
f 9 h. «	7*	9 ‘	K
6.	Let a, 5, <?, be the four magnitudes.
Λ C	Cl^
Dem.—a : b : : c : d, .·. ^	.*. ^ = j2 ; that is, α* :
Similarly a3 : 53 : : c3 : d3.
7.	Let a, b, c, d; a, b> c, d', be the two sets of magnitudes.
It is required to prove that d = d'.
*	a	c	λ	c
Dem.—a	z	b 1	: e	1 d,	and	a :	b :	: e :	d\	.·.	7 =	3, and - =	—;
O d	9 m
.·. 3	^. Hence d = (Γ.
d d
8.	Dem.—Since the three magnitudes are continual proportions,
we have \ and -■=■-. Multiplying these equalities, we get
be c c
ti	b2	ct	b	/ (l	\
-	=	— ;	that	is,	a :	c	:	:	b2	:	c*.	Again,	- = -;	.·.	( — —	11
e	<r	o	e	\o	/
/ b \	a — b b — c	(a - b)2	(b — e)2
\c /	b	c	b-	<r
that is, {a — b)2 : (b — <r)2::	: c2. Hence we have a : c :: (a— b)*
:(b-e)2.
9.	Dem.—AC : CB : : AD : DB (hyp.), .·. AC - CB : AC
B O'
D
+ CB : : AD - DB : AD + DB; that is, 2OC : 2OB : : 20B
: 2 OD. Hence OC : OB : : OB : OD.186
EXERCISES ON EUCLID.
[book V.
10.	Dem.—Because CD is bisected in O', and produced to 0,
we have (II. vi.) OD.OC + O'C2 = OO'2; but OD.OC = OB*
(Ex. 9); .*. OB2 + O'C2 = OO'2; that is, OO'2 = OB2 + O'D2.
11.	Dem.—AC : CB : : AD : DB (hyp.), .*. AC : AB — AC
AP	AD
” iD ’ “ -4B· ·■· 4B = AC * AD^AB' ··■“<“- AI>
= AD (AB-AC); .·. AC.AD - AC.AB = AD.AB - AD.AC.
A________"
Transposing, we get 2 AC.AD = AB (AC + AD). Divide by
2 11
AB.AC.AD, and we have -τ^ζ = tt=- + —.
1	AB AD AC
12.	Dem.—BD : BC : : AD : AC (hyp.). Working, as in
Ez. 11, we get A =	.
13.	Dem.—AC : CB :: AD : BD (hyp.), .·. AC.BD = CB.AD,
.·. AC.BD + CB.AD = 2 CB.AD. Again, AB.CD = (AC
+ CB) (CB + BD) = AC. BD + AC. CB + CB2 + CB. BD
= AC . BD + CB (AC + CB + BD) = AC. BD + CB.AD
= 2 CB.AD.BOOK ΤΙ.]
EXERCISES ON EUCLID.
18Γ
BOOK YI.
PROPOSITION II.
1.	Let AE, BE be two lines cut by the parallels AB, CD, EF.
It is required to prove that AC : CE : : BD : DF.
Dem.—Join BE, cutting CD in G. Now in the Δ AEB we
have AC : CE : : BG : GE (n.); and in the Δ BEF, BG : GE
: : BD : DF. Hence AC : CE : : BD : DF.
PROPOSITION III.
2.	(1) Dem.—Through C draw CF || to AD. Now (I. xxix.)*
the L EAD = EFC, and the L DAC = ACF; but EAD = DAC
(hyp.); .·. AFC = ACF; .*. AC = AF. Now (ii.) BA : AF
: : BD : DC ; that is, BA : AC : : BD : DC.
(2) Dem.—Produce AB through A, cut off AE = AC, and
join DE. Now (I. iv.) the Δ8ΕΑϋ, CAD are congruent; .*. DE
= DC, and the L ADE = ADC; hence BD : DE : : BA : AE
that is, BD : DC : : BA : AC.
3.	Let AB be the base, ACB the vertical L, CO, CO' the in-
ternal and external bisectors. It is required to prove that AB is
divided harmonically in 0 and O'.
Dem.—In the Δ ACB, AO : OB : : AC : CB (iii.) ; and in the
same Δ, AO': O'B : : AC : CB (Ex. 1); .·. AO : OB : : AO': O'B.
Hence AB is divided harmonically in 0 and O'.
4.	This is the same as Ex. 3.
5.	Let ACB be the right L, AB the line intersecting the sides
CA, CB, and CD, CE any two lines making equal L · with CA..
Produce BA to meet CD. It is required to prove that AB is cut
harmonically in E and D.
Dem.^-Produce DC to F. Now in the Δ DCE, DA : AE
: : DC : CE (hi.). Again, since the L ACB is right, the L *
DCA, BCF are together equal to a right L; but DCA = ACE;
··· ECB = BCF ; BC is the bisector of the external L ECF ;18$
KXERCISES ON EUCLID
[book yi.
lienee (Ex. 1) DC : CE : : DB : BE ; .*. DA : AE : : DB : BE*
Hence AB is cut baimonically in E and D.
6. Let AB be the base, AC and CB the sides.
Sol.—Bisect the L ACB by CD. Produce AC to F, and bisect
the L BCF by CE, meeting AB produced in E.
Now AD : DB : : AC : CB (hi.) ; but the ratio AC : CB is given
{hyp.); .·. the ratio AD : DB is given; D is a given point.
Again, AC : CB : : AE : EB (Ex. 1); .·. the ratio AE : EB is
.given, and AB is given; hence the point E is given. And
because the L ACD = BCD, and FCE = BCE, the L DCE is
right; hence the Θ on DE as diameter will pass through C;
and because the points D, E are given, it will be a given Θ.
It divides the base in the points D, E harmonically, in the ratio
of AC : CB, and is the locus of the vertex. It is called the
u Apoll·»nian locus.”
7. Bern.—b : c :: CD : DB (in.); .·. b + e : e : : CD + DB
^ DB ; but CD + DB = CB = a; .·. b -f e : c : : a : DB;
(b -f c) DB = ac; hence DB =
b + c
....	. ^ ae ae 2 abc
Adding, we get DD' = —— + 7— = rx—
b + c b —c b2- c1
Similarly, D'B =
b-cBOOK VI.]
EXERCISES ON EUCLID.
189
8. (1) Dem.—In the last Exercise we got DD' =
1
• ' DD'
b2-c2
2 abc ’
1 c2 —
Similarly, =v,, - -
1 EE’	2 abc
.Adding, we get A-. + ~ + ^ = 0.
»andFF =
2 'abc
a» - y
2 abc
(2) Dem.—From (1) we have
1 b2 — c2 a2 a2b2 - c2a2
DD' 2abc * *# DD' 2 abc *
Similarly,
b2 b2c2-a2b2	c2 c2a2 — b2&
EE7- '2 abc ,andFF;“ 2 abc *
Adding, we have
a2 b2 c2
DD7 + EE' + FF' = '
PROPOSITION IV.
1. Dem.—Let 0, O' be the centres. Join OA, O'A', and let
fall _L8 OC, O'C' on AA\ From T let fall a _L TD on AAr.
Now in the Δ8 ACO, ADT we have the L8 ACO, ADT equal,
and the L OAT is right (III. xvm.), and is equal to the sum Jof
the L8 OAC, AOC. Reject OAC, and we have AOC = TAD
T
the Δ· OAC, ADT are equiangular; hence (iv.) OA : AC
AT : TD; alternation, OA : AT :: AC : TD. Similarly,
O'A' : A'T :: A'C': TD; hut AC = A'C', since AB, A'B' are190
EXERCISES ON EUCLID.
[book VI.
equal, and (III. hi.) are bisected in C, C'; hence the ratio of
AC : TD is equal to the ratio of A'C' : TD; .·. AO : AT
: : ΑΌ' : A'T; alternation, AO : ΑΌ' : : AT : A'T.
2.	Dem.—As in the last exercise, let AB, A'B' be the chords,
having a given ratio. The construction will be as before, and
we will have the Δ8 ATD, ACO equiangular; then (iv.) AT : TD
: : AO : AC, and TD : TA' :: C'A' : ΑΌ'. Multiply together,
and we get AT : TA' :: AO . C'A' : AC . A'O'; hence
AT _ AD GA' _ AO AB'
* TA ” AO' ‘ AC".“ AO' ’ AB *
3.	Let ABC be the Θ, and DE the line.
Sol.—From the centre 0 let fall a _L OF on DE. Draw PG
a tangent. In FO take FH = FG. H is the required point.
Dem.—Through H draw CD, meeting the Θ in B, C, and DE
in D. From B, C let fall ±8 BJ, CE on DE, and draw AD a
tangent to the circle.
Now (“ Sequel,” Book III., Prop, xxi.) DA2 = DF2 + PG*
= DF2 + FH2 = DH2; but DA2 = CD . DB (III. xxxvi.);
.·. CD . DB = DH2; .·. CD : DH : : DH : DB.
Again, CD : DH : : CE : HF (iv.), and DH : DB:: HF : BJ;
.·. CE : HF :: HF : BJ ; that is, CE . BJ =HF2.
4.	Let the given ratio be the ratio of AC : AG.
Sol.—Join BD, and produce GA to H, so that GH . AH = BD2
(II. vi., Ex. 2). Place a chord BF in ADB = AH (IV. i.).
Join AF. AF is the required line.
Dem.—Join AD, and produce CD, BF to meet in J.
Now the L AFB is right (III. xxxi.), .*. AFJ is right, and
AC J is right; hence ACFJ is a cyclic quadrilateral; hence JB « BF
= AB. BC; but AB. BC = BD2, thatis = toGH . AH; !.*. JB· BF
= GH. AH; but BF = AH (const.); .·. JB = GH, and JF = AG-191
book vi.] EXERCISES ON EUCLID.
Again, AC:CE : : JF : EF (iv.); alternation, AC : JF :: CE
; EF; but JF = AG. Hence AC : AG : : CE : EF.
PROPOSITION X.
3.	See “ Sequel,” Book VI., Prop, v., Sect. iii.
4.	Let the sum of the squares of the lines be equal to the
squares on AB, and their ratio that of mm.
Sol.—On AB as diameter describe a Θ ABC. Divide AB in
D, so that AD : DB : : m : n (Ex. 1). Bisect the arc ACB in C
Join CD, and produce it to meet the circumference in E. Join
AE, BE. AE, BE are the required lines.
Dem.—AB2 = AE2 + BE2, and (III. xxvit.) the L AEB is
bisected; hence AE : EB :: AD : DB ; but AD : DB : i min.
Hence AE : EB i \ m i n.
5.	Let the difference of the squares of the lines be equal to
AB2, and’their ratio that of m i n.
Sol.—Divide AB internally and externally in C and D, in the
ratio of tn i n. On CD as diameter describe a semicircle. Let192	EXERCISES ON EUCLID. [book ti.
O be its centre. Erect BE X to AD, and join AE. AE and BE
are the required lines.
A C B O	D
Bern.—Join OE, CE. Now (I. xlvii.) AE2 — BE2 = AB2. And
because AB is divided harmonically in C and D, and CD is bisected
in 0, OB, OC, OA are in geometrical progression (Book V., Ex. 9).
Hence OA . OB = OC* = OE2, the L AEO is right, .·. the
L OAE = BEO ; but ECO = CEO (I. v.). Hence (I. xxxn.)
the L AEC = CEB ; .·. (in.) AE : EB : : AC : CB; that is,
: : m : n.
6. (1) Let AB be the base; m : n the ratio of the sides, and
the rectangle X the area.
Sol.—Divide‘AB internally and externally in C and D, in the
ratio m : n (Ex. 1). On CD as diameter describe a Θ ; to AB
apply a parallelogram AF, whose area is 2 X. Let its side EF
cut the Θ in G. Join AG·, BG. AGB is the Δ required.
Dem.—AG : GB : ; AC : CB (Dem. of Ex. 4), that is as
m : n, and the parallelogram AF = 2 AGB; but AF = 2 X;
.·. AGB = X.
(2) Let AB be the base; min the ratio of the sides, and E*
the difference of the squares of the sides.
Sol.—Divide AB as in (1). On CD as diameter describe a Θ.
Divide AB in F, so that AF* - BF2 = R\(“ Sequel,” Book I.,BOOK VI.]
EXERCISES ON EUCLID,
193
' Prop. ix.). Erect FE X to AD, cutting the Θ in E. Join AE,
BE. AEB is the Δ required.
Bern.—AE : EB :: AC : CB, that is as m : n, and AE2 - EB2
= AF2 - FB2 = R2.
(3)	Let AB be the base ; m : n the ratio of the sides, and 2 R2
the sum of the squares of the sides.
Sol.—Divide AB as in (1), and on CD as diameter describe a
Θ CDE. Bisect AB in F. Erect FG X to AD. From A inflect
AG on FG, and equal to R. With F as centre, and FG as
radius, describe a Θ, cutting CDE in E. Join AE, BE. AEB
is the Δ required.
Bern.—Join FE. Now as in (1) AE : BE :: m : «, and
FG = FE (const.); FG2 = FE2; AF2 + FG2; that is, AG2
= AF2 + FE2, .-.2 AG2, that is, 2 R2 = 2 AF2 + 2 FE2. Hence
(II. x., Ex. 2) AE2 + BE2 » 2 R2.
(4)	Let AB be the base; m:n the ratio of the sides, andJX
the vertical L.
Sol.—Divide AB as in (1). On CD as diameter describe a Θ
CDE; and on AB describe a Θ AEB, containing^an L = X.
Join AE, BE. AEB is the Δ required.
Bern.—AE : BE : : m : n, and the vertical L AEB ='X.
(5)	Let X be the difference of the base angles.
Sol.—Divide AB as in (1), and on CD describe a Θ CDF*
Erect CE X to AD ; and at C, in the line CE, make the L ECF
= J X. Join AF, BF. AFB is the Δ required.
Bern.—AF : BF :: m : n; and the difference between the
L · ACF, BCF is equal to 2 ECF = X; but ACF = CBF + CFB,
and BCF = CAF + CFA, and CFA = CFB. Hence CBF
- CAF = ACF - BCF = X.194
EXERCISES ON EUCLID.
[book VI.
PROPOSITION XI.
1.	Dem.—Join OB, B'C, &c. Now in the Δ8 OAB, BB'C, we
have OA : AB : : B'B : BC, and the right L OAB = B'BC ; hence
(vi.) the Δs are equiangular, . ·. the L ABO = BCB'; hence OB, B'C
are parallel. Similarly B'C, C'D are parallel. Now, since the
lines AO, BB', CC' are parallel, we have (n., Ex. 1) OB' : B'C'
: : AB : BC; and because OB, B'C, C'D are parallel, OB' : BC
: : BC : CD ; hence AB : BC : : BC : CD. In like manner
BC : CD : : CD : DE. Hence AB, BC, CD, &c., are in con-
tinued proportion.
2.	Dem.—Because B'M is || to Ail, the Δ8 OMB', OAfl are
equiangular; .*. OM : MB' : : OA : Ail; hut OM = OA — AM
= AB - BB' = AB - BC, and MB' = AB, and OA = AB. Hence
AB-BC : AB: : AB : Ail.
PROPOSITION XIII.
1.	(Diagram to Prop. vm.).
Sol.—Let AB, BD be the two lines. On AB describe a semi-
circle. At D erect DC ± to AB, and meeting the semicircle in
C. Join BC. BC is a mean proportional between AB, BD.
Dem.—Join AC. Now the Z.SABC, BCD are equiangular
(viii.), .*. AB : BC : : BC : BD. Hence BC is a mean pro-
portional between AB and BD.
2.	Sol.—Let 0 be any point taken within a Θ ABC ; O' the
centre. Join 00', and produce both ways to meet the circum-
ference in A, B. Through 0 draw CD JL to AB. CD is bisected
in 0 (III. hi.). Through 0 draw any other chord FE. OC is
a mean proportional between OF and OE.
Dem.—Join CF, DE. Now, because the Δ® OCF, OED are
equiangular (III. xxi.), we have (iv.) OF : OC : : OD : OE;
but OD = OC; .'. OF : OC : : OC : OE. Hence OC is a
mean proportional between OF and OE.
3.	Let ABC be a Θ, 0 any external point. From B draw a
secant BO, and from 0 draw OC a tangent to the Θ. It is
O■HOOK VI.]
EXERCISES ON EUCLID.
195
required to prove that OC is a mean proportional between OB
and OA.
Dem.—Join AC, BC. Now in the Δ8 OAC, OBC, we have
the L OCA = OBC (III. xxxii.), and the L BOC common;
hence the Δδ are equiangular, .·. BO : OC : : OC : OA. Hence
OC is a mean proportional between OB and OA.
4.	Dem.—Let AB be the chord of the arc. Join AE, AC, CB.
Now because the arc AC = BC, the L CAB = CBA; but CBA =
AEC (III. xxi.) ; .'. AEC = CAD, and the L ACD is common ;
the Δ8 ACE, ACD are equiangular; .*. EC : AC : : AC : CD.
Hence AC is a mean proportional between CE and CD.
5.	Let ACB be a Θ whose diameter is AB ; FG, HJ two
parallel chords; CDE a Θ touching ACB internally in C; and
FG, HJ in D, E. From 0, the centre of CDE, let fall a _L OK
•on AB.’ It is required to prove that OK is a mean proportional
between AG and JB.
Dem.—Join OD, OE, CD, CE. CD, CE produced must
o 2196
EXERCISES ON EUCLID.
[book VI*
pass through A, B (hi., Ex. 51). Now (III. xviii.) the L ODG
is right, and DGB is right; .·. OD is || to AB. Similarly OE is
|| to AB ; .*. OD, OE are in the same straight line. Again,
since the L AGD is right, the L8 GAD, GDA are equal to a
right L ; and because ACB is right (III. xxxi.), the L8 CAB,
CBA are equal to a right L; hence the L GDA = JBE, and the-
L DGA = EJB; .·. the Δ8 ADG, JEB are equiangular; hence
AG : GD :: EJ : JB ; hut GD and EJ are each equal to OK ;
.·. AG : OK : : OK : JB. Hence OK is a mean proportional
between AG and JB.
6. Let ADB be a semicircle whose diameter is AB; CEF a Θ
touching ADB in E and AB in C. Through 0, its centre, draw
the diameter CF, and produce it to meet ADB in D. It is
required to prove that CF is a harmonic mean between AC
and CB.
Dem.—AB.r = CD2 (“Sequel,” III., Prop, v.); hut AC.CR
Hence (V., Miscellaneous, Ex. 11.) 2 r, that is CF, is a harmonic
mean between AC and CB.
7. Let ACB he a Θ whose diameter is AB; FG, HJ, two
parallel chords meeting the Θ in F, H, and the diameter in G, J.
Describe a Θ CDE touching ACB externally in C, and GF, JH
produced in D, E. From 0, its centre, let fall a _L OK on AB.
It is required to prove that OK is a mean proportional between
AJ and GB.
The proof is the same as in Ex. 5.KOOK VI.]
EXERCISES ON EUCLID.
197
PROPOSITION XVII.
2. Dem.—Describe a Θ about the Δ. Produce AC to G, and
bisect the external L BCG by CD', meeting AB produced in D'.
Produce D'C to meet the Θ in E, and join AF. Now the
L BCD' = GCD', and GCD' = FCA, .·. BCD' = FCA; and
since the Zs CBD', CBA are together equal to two right Z.s,
.and the L8 CFA, CBA are equal to two right L8, the L CBD'
= CFA; .·. the Δ8 AFC, BCD' are equiangular; .·. AC : CF
* : D'C : CB (iv.); hence AC . CB = D'C . CF. Again AD'. D'B
= FD'.D'C; but FD'. D'C = FC. CD' + CD'2 (II. iii.) = AC. CB
4- CD'2. Hence AD'. D'B' - CD'2 = AC . CB.
4. Dem.—Let 0' be the centre of the escribed Θ, touching
ΛΒ externally, and the other sides produced. Join O'C, cutting
1 he circumscribed Θ in E. Through E draw EF, the diameter
<of the circumscribed Θ. Join O'B, EB, FB, O'G, G being the
point where CB produced touches the escribed Θ.
Now the Z.8 O'GC, EBF are equal, each being right, and the
Z.O'CG = EFB (III. xxi.) ; .·. the Δ8 O'CG, BFE are equi-
angular ; hence (iv.) FE : EB :: O'C : O'G, and EB = EO' (Dem.
Of iv., Ex. 19); hence FE : EO' :: O'C : O'G, .·. FE . O'G
= EO'. O'C ; that is, the rectangle contained by the diameter of
the circumscribed Θ, and the radius of the escribed Θ, is equal
to the rectangle contained by the segments of any chord of the
•circumscribed Θ passing through the centre of the escribed Θ.
7. Dem.—Produce AD to meet the circumference in G; then
(Ex. 6) we have AB . AE + AC . AF = AG . AD ; but AG . AD
= GD . DA + DA2 (II. iii.), and GD . DA = BD . DC (III.
xxxv.). Hence AB . AE + AC . AF = BD . DC + DA2.
9. Dem.—Let ABC be the Δ, and FGG'F' the inscribed
square. From B let fall a J_ BD on AC, cutting the side FG
of the 0quare in E.
Now AC : FG :: AB : FB (iv.) ; but AB : FB : : BD : BE
(iv.); .*. AC : FG :: BD : BE. Hence, putting b for base, p for198
EXERCISES ON EUCLID.
[book VIw
X, and 8 for side of square, we have b : s :: p : p — s; .·.
bp — bs = sp, Hence bp — (b + p) s.
B
10. Dem.—Let ABC be the Δ, and FGF'G' the escribed
G	K F
square. From B let fall a X BD on AC, and produce it to meet
FGinE.
Now AC : FG : : AB : BF (iy.) ; but AB : BF : : BD : BE
(iy.) ; AC : FG :: BD : BE; that is, putting s’ for the side
of the square, b : 8* : : p : $' —p. Hence bsr — bp — $rp\ .·. bp
= s {b-p).
11. From P let fall a 1 PC on tbe chord AB, and from A, B
let fall Xs AD, BE on DE, the tangent at P. It is required to·
prove that CP* = AD . BE.
Dem.—Join AP, BP. Now in the Δ8 APD, BPC, the L API>
= BPC (III. xxxii.), and the L ADP = BCP, .*. the Δ* are
equiangular; hence (iv.) AP : AD : : BP : PC; alternation,
AP : BP :: AD : PC. In like manner for the Δ8 APC, BPE„BOOK VI. J
EXERCISES ON EUCLID.
199
we have AP : BP : : PC : BE, .·. AD : PC : : PC : BE. Hence
CP* = AD . BE.
12. Dem.—In the Δ8 AOD, BOG, the L AOD = BOC, and
the L 0AD = OBC (III. xxi.); hence (iv.) AD : AO: : BC: BO;
alternation, AD : BC :: AO : BO. Multiplying each by AB, we
get AD . AB : AB. BC :: AO : BO. Similarly AB. BC : BC . CD
: : BO : CO, &c. Hence the four rectangles are proportional
to the four lines.
14. Dem.—Draw the diagonals AC, BD. Make the L ABO
= DBG, and BAO = BDC. Join OC.200
EXERCISES ON EUCLID.
[book VI.
Now the Δ8 ABO, DBC are equiangular, .·. AB : AO : : BD
: DC,.AB. CD = AO . BD. Again, since AB : BO :: Bp : BC ;
alternation, AB : BD :: BO : BC; and since the L ABO = DBC,
the L ABD = OBC; hence (iv.) the Δ® ABD, OBC are equi-
angular ; . ·. (it.) AD : BD :: OC : BC; hence AD . BC = BD . OC.
Now we have proved AB. CD = AO . BD, AD . BC = OC . BD,
and AC . BD = AC . BD ; hence the three rectangles are propor-
tional to the sides AO, OC, AC of the Δ AOC; and since the
Δ8 AOB, CDB have been shown to be equiangular, the L AOB
= BCD; and because the Δ8 BOC, ABD are equiangular, the
L COB = BAD. Hence the L AOC is equal to the sum of the
L8 BAD, BCD.
15.	Let ABCD be a cyclic quadrilateral; AC, BD its diagonals.
At P, any point in the circumference of the circumscribed Θ, draw
a tangent to the Θ, and let fall Xs PE, PF, PG, PL on AB, BD,
AC, CD. It is required to prove that PF . PG = PE . PL.
Dexn.—From A, B, C, D let fall Xs AH, BI, CJ, DK on the
tangent at P. Now PF2 = BI. DK (Ex. 11), and PG2 = AH . C J ;
.·. PF2 . PG2 = BI . DK . AH . CJ. In like manner PE2 . PL2
= BI. AH . DK. CJ; .% PF2. PG2 = PE2 . PL2. Hence PF. PG
= PE. PL.
16.	Dem.—The L APB is right (III. xxxi.); .*· DPE is
right, and equal to ECB, and PED = CEB, .·. PDE = CBE.
Now since PDE = CBE, and ACD = ECB, the Δ8 ADC, EBC
are equiangular; hence AC : CD :: CE : CB (iv.), AC . CB
= CD . CE; but AC . CB = CF2 (xvii.) ; .·. CD . CE = CF2.
Hence CF is a mean proportional between CD and CE.book vi.] EXERCISES ON EUCLID.
201
PROPOSITION XIX.
1.	Let ABC, DEF be the two Δ8. Now AB = f DE (hyp.),
.·. AB : DE :: 3 : 2 ; .·. AB2 : DE2 : : 9 : 4 ; but ABC : DEF
:: AB2 : DE2 (xix.) Hence the Δ ABC : DEF :: 9 : 4.
2.	Let AB be a side of the inscribed polygon, O the centre of
the Θ. Join OA, OB, and bisect the L AOB by OP', meeting
AB in P. Through P' draw a tangent to the O, and produce
OA, OB to meet it; then evidently A'B' is a side of the circum-
scribed polygon.
Now, if each of the polygons have n sides, and we denote their
ΊΓ	7t'
areas by ir and π, we have the Δ AOB = and A'OB' = — ;
J	9	η	n
hence AOB : ΑΌΒ':: ir : n·'; but (xix.) AOB : A'OB' :: AO2
: A'O2 ; that is, : : OP2 : OP'2 (iv.), or: : OP2: OA2 ; hence x: »'
:: OP2 : OA2;	x : ir : : AP2 : OA2; that is, as 4 AP2
: 4 OA2; that is, as AB2 is to the square of the diameter; but w
is less than the square of the diameter (iv., Ex. 37). Hence ί/ — v
is less than AB2.
PROPOSITION XX.
4. Dem. —Let AB, BC, CA be three given lines in the form,
of a Δ. Inscribe in ABC a Δ A'B'C' similar to the Δ FDE.
About the Δ8 A'BC', A'B'C describe Θ8 intersecting in 0 ; then202
EXERCISES ON EUCLID.
[BOOK VI.
the Θ about AB'C' will pass through 0 (hi., Ex. 28). Join OA',
OB, OC, OB', OC', AA'. Now (III. xxi.) the L BOA' *= BC'A',
and COA' = CB'A'; .*. the L BOC is equal to the sum of the
L · BC'A', CB'A'; hut BC'A' = BAA' + AA'C', and CB'A' = CAA'
+ AA'B'; the L BOC = C'AB' + C'A'B'; hut C'A'B'= FDE;
hence BOC = C'AB' -f FDE; hut the L FDE is given, and C'AB'
is given; the L BOC is given, and the base BC is given ;
A
hence the O described about the Δ BOC is given in position.
Similarly, the Θ5 * * 8 about the Δ8 AOB, AOC are given in posi-
tion ; hence 0 is a given point. Hence, if we inscribe another
Δ A"B"C" similar to FDE in ABC, the Θ8 described about the
Δ8 A"BC", B"CA", C"AB" will co-intersect in 0, and if we join
the angular points to 0, the L8 OC''A", OA"C" will be equal to
the L8 OBA', OBC'; that is, equal to the L8 OC'A', OA'C; hence
the Δ8 OC'A', OC"A" are equiangular, and therefore (Ex. 2) O
is the centre of similitude of the Δ8 A'B'C', A"B"C".
5. Let ABODE, A'B'C'D'E' be two similar figures, having the
sides AB, BC parallel to the sides A'B', B'C'. It is required to
prove that the other homologous sides are parallel.
Dem.—Join AA', BB', and produce them to meet in F. Now
the L BAF= B'A'F (I. xxix.); but since the figures are similar,,
the L BAE = B'A'E'; hence the L FAE = FATS', and thereforeKOOK VI.]
EXERCISES ON EUCLID.
20$
the line AE is parallel to ΑΈ'. Similarly, it can be shown that
the other homologous sides are parallel.
6. Let ABC, DEF be the homothetic figures. Join BE, ADr
and produce them to meet in G. Join CF. It is required to
prove that CF produced will pass through G.204
EXERCISES ON EUCLID.
[book yi.
Dem.—If not, let it pass through H. Produce AG to H.
Now the L GED = GBA (I. xxix.), and the L GDE>= GAB ;
hence (iy.) AG : AB : : DG : DE ; hut AB : AC :: DE : DF;
AG : AC : : DG : DF ; alternation, AG : DG : : AC : DF.
Again, since the Δ8 HAC, HDF are equiangular, we have
AH : AC : : DH : DF; alternation, AH : DH : : AC : DF;
-·. AH : DH : : AG : DG; hence (V. xvn.) AD : DH : : AD
: DG; and therefore DH = DG, which is absurd. Hence CF
produced must pass through G.
7. Dem.—Let ABC, DEF he the two similar figures; 0 their
centre of similitude. Join OA, OB, OC, OD, OE, OF. From
OA, OB, OC cut off OD', OE', OF' equal respectively to OD,
OE, OF, and join D'E', D'F', E'F'. Now since OD' = OD,
OE’ = OE, and the L DOE' = DOE (hyp.), .·. DE = D'E',
.and the L OED = OE'D'; hut OED = OBA (hyp.); .*. OE'D'
= OBA; .*. D'E' and AB are parallel. Similarly, D'F is || to
AC and equal to DF, and E'F is equal to EF and || to BC;
hence the figure DEF may he turned round 0 so as to take
up the position D'E'F7. In like manner the figure may he turned
round in the opposite direction, as in the second diagram.
10. Dem.—Let 0, O' he the centres of the Θ8, and A one of
their centres of similitude. Join 00', and from A draw AB, AC,
AD, AE tangents to the Θ8. Join OA, OB, O'A, O'D.
Now since A is a centre of similitude, the L BAC = DAE;
therefore their halves are equal; that is, the L BAO = DA0%BOOK VI.]
EXERCISES ON EUCLID.
205*
and the right L8 ABO, ADO' are equal; .*. the Δ8 ABO, ADO'
are equiangular; hence AO : OB : : AO' : O'D; alternation,
AO : AO':: OB : O'D; but the ratio OB : O'D is given, since
OB and,0'D are given lines; hence the ratio AO : A'O is given..
Now in the Δ OAO' we have the base 00' given, and the ratio
of the sides. Therefore (in., Ex. 6) the locus of A is a circle.
PROPOSITION XXI.
1. Dem.—Let AA', BB', CC' be corresponding sides of the
similar rectilineal figures ; then since the figures are homothetic,
these sides are parallel. Join BA, B'A', and produce to meet
in 7; then because AA', BB' are corresponding sides of the206
EXERCISES ON EUCLID. [book vi.
homothetic figures, y will be their centre of similitude. In
like manner, if we join BC, B'C', and produce to meet in a;
AC, A' C' to meet in β ; a and β will be centres of similitude.
AA'-
Now (IY.)	:
7A
Similarly, g = BB„
Λ AjS AA'
and*c =
,	,	. BB' CC' AA'.	B-y
but the product of —·, Tr—, is unity; .·. the product of —f,
AA ±>±> tt	<yA
—, ~ is unity. And hence (“ Sequel,” Book VI., Prop, iy.,
ax) /SO
Cor. 1, p. 69), the points a, β, y are collinear.
PROPOSITION XXIII.
1.	Dem.—Let ABC, DEF be the Δ8 having the Z ABC
= DEF. Complete the parallelograms ABCG, DEFH. Now
the Δ ABC : DEF : : ABCG : DEFH; but ABCG : DEFH
: : AB . BC : DE . EF (xxm.). Hence ABC : DEF : : AB . BC
: DE. EF.
2.	Let ABCD, EFGH be two quadrilaterals whose diagonals
AC, BD ; EG, FH intersect in I, J, making the Z CIB = GJF.
It is required to prove that ABCD : EFGH :: AC . BD : EG. FH.
Dem.—The area of ABCD is equal to the area of a Δ having
two sides equal to AC, BD, and the contained Z equal to CIB
(I. xxxiv., Ex. 7); and EFGH is equal to a Δ having two sides
equal to EG, FH, and the contained Z equal to GJF; but (Ex. 1)
those Δs are to one another as AC. BD : EG. FH. Hence ABCD
: EFGH :: AC.BD : EG.FH.
PROPOSITION XXX.
1. Let ABC be a right-angled Δ whose sides are in continued
proportion; that is, having AB : BC :: BC : CA. From C let
fall a X CD on AB. It is required to prove that AB is divided
in extreme and mean ratio in D.
Dem.—Because AB : BC :: BC : CA, AB.AC = BC2. Again
(I. xlvii., Ex. 1), AB.BD = BC2; .·. AC = BD, and AB.AD
= AC2; AB.AD = BD2. Hence AB is divided in extreme
•and mean ratio in D.BOOK VI.]
EXERCISES ON EUCLID.
207
2.	See Demonstration of last Exercise.
3.	Join FD, and describe a Θ about the Δ FHD. Let 0 be
its centre. Join DO, and produce it to meet thejcircumference
in I. It is required to prove that DI2 = 6 FD2.
Dem.—Join IF. Produce FH, and let fall a _L DJ on it.
Now since AGr is a square, AF = AH ; .*. the L AHF = AFH,
and FAH is a right L; .·. AHF is half a right L; .·. BHL is
half a right L, and HBL is a right L; . ·. HLB is half a right L,
and BH = BL, and the L DLJ = BHL; .·. DLJ is half a right L,
and DJL is a right L; .·. JDL is half a right L ,*and JL = JD;
.·. JL2 = JD2, and DL2 = 2DJ2.
Again, since AB = DB, and BH = BL, .*. DL^p AH; but AB
is divided in extreme and mean ratio in H, .·. BD is divided in
extreme and mean ratio in L; and hence (II. xi., Ex. 4) BD2
+ BL2 = 3 DL2 = 6 DJ2; hence BD2 + ΒΉ.2; that is, DH2 =
6 DJ2. Again (III. xxn.), the Ls FHD, FID are together equal
to two right /.*, and the L8 FHD, DHJ are equal to two
right ZB; the L FID = DHJ, and the right L IFD = HJD;
the Δ8 IFD, DHJ are equiangular; .·. ID : DF : : DH :
DJ; .·. ID2 : DF2: : DH2 : DJ2; but DH2 = 6DJ2. Hence ID*
= 6DF2.208
EXERCISES ON EUCLID.
[book yi.
PROPOSITION XXXI.
1. Dem.—Let ABC be the semicircle, of which AB, CB are
supplemental chords. On AB, CB describe semicircles ADB,
BEC. Now (xxxi.) the semicircle ABC is equal to the sum of
the semicircles ADB, BEC. Take away the common segment»
AFB, BGC, and we have the Δ ABC equal to the sum of the·
crescents ADBF, BECG.
Exercises on Book VI.
1.	Let ACB be a fixed Δ, DE a parallel to AB. Draw the
diagonals AE, BD, intersecting in 0. Join CO, and produce it
to meet AB in H. It is required to prove that CH bisects AB.
Dem.—Through 0 draw FG parallel to AB. Now (II.)
AE : EO :: BD : DO; but, by similar Δβ, AE: EO : : AB : OGr
and BD : DO : : AB : OF ; hence AB : OG : : AB : OF, and
therefore OG = OF. Now ACB is a Δ ; and FG, a parallel to
the base, is bisected by CO. Hence AB is bisected by CO.
2.	Let 0 be the centre of the Θ, and P the given point.
From P draw PA to any point A in the Θ. Divide AP at B in
a given ratio. It is required to find the locus of B.
Sol.—Join OP, OA, and draw BC || to AO.BOOK VI.]
EXERCISES ON EUCLID.
209
Now PB : BA : : PC : CO (n.); but the ratio PB : BA is
given; PC : CO is given, and therefore C is a given point.
Again, by similar Δ ·, we have PA: AO :: PB : BC; alternation,
PA : PB:: AO: BC ; but the ratio PA : PB is given, .·. AO: BC
is given; but AO is given; .*. BC is given, and the point C is
given. Hence the locus of B is a Θ, having C as centre and
BC as radius.
3.	Dem.—Through B, C draw BE, CD || to XT.
Now, by similar Δ8, AC : AD : : CB : CE; alternation, AC
: CB : : AD : CE, .*. AD : CE : : m : n; but AD = AA' - A'D
= AA' - CC', and CE = CC' - C'E = CC' - BB; hence AA'
— CC': CC' — BB' : : m : n; .·. n AA' -nCC'=m CC' - i»BBr;
and hence m BB' + n AA' = (m + n) CC'.
4.	See “ Sequel,” Book VI., Prop, ii., Section 1.
5.	See “ Sequel,” Book VI., Prop, iv., Section 1.
6.	Dem.—Let the rectangle AB . AC = k2. Produce AB to
meet the circumference in D. Now, if t denote the tangent drawn
rom A to the Θ (III. xxxvi.), AB . AD = &; .·. AB . AD :
AB . AC : : t2 : A2; that is, AD : AC ::<*:**; but the ratio
f*: k2 is given, AD : AC in a given ratio, and hence (Ex. 2)
he locus of C is a circle.210
EXERCISES ON EUCLID.
[book VI.
7.	Dem. — Join 0, the centre of the inscribed Θ, to the
points F, G, H, where the sides AB, AC, BO touch the 0.
Join OC.
Now since AF = AG, BF = BH, and CG = CH, .·. AB — AC
= BF - CG = BH - CH = 2DH. Again, AB2 - AC2 = BE2
- EC3 (I. xlvii.) ; that is (AB + AC) (AB - AC) = (BE
+ EC) (BE - EC); .*. (AB + AC) 2DH = BC.2DE; hence
(AB + AC): BC :: DE : DH. Again (hi.) AB : AC :: BL : LC;
.·. (AB + AC) : AC :: BC : LC; .·. (AB + AC): BC :: AC : LC.
Again, AC : LC :: AO : OL (in.); but AO : OL :: TIE : ITT,
(n.), AC : LC :: HE : HL; hence (AB + AC) : BC :: HE
: HL; that is, DE : DH :: HE : HL; and hence DE.HL
= HE. HD.
8.	Dem.—Let O' be the centre of the escribed Θ, touching
BC produced in K. Now (AB + AC) : BC :: AC : LC (Ex. 7);
that is, as AO : OL, .·. (AB + AC + BC) : BC :: AL : OL :: LE
: LH; .*.2 BE : 2 BD :: LE : LH ; hence LH. BE = BD. LTP)l
9.	See Book VI., Prop, xvii., Exs. 3, 4.
10.	Dem.—From Ex. 9 we have d2 — R2 — 2 Er; d* = Έρ
+ 2Rr'; rf"2 = R2 + 2 Rr", and rf"2 = R2 + 2Rr"'; .·. d2 + d'2
+ d"2 + d,n2 = 4 R2 + 2 R (r' + r" + r"' - r); but (Book III.
Ex. 19) (/ + /' -f — r) = 4 R. Hence d2 + d'2 + d"* + #
= 4R2+2R.4R = 12 R2.
11.	(1) Dem.—Let the sides of the Δ be denoted by a, by c.
Now (IV. iv., Ex. 9) rs = Δ; s = Again, ap' = 2Δ
(II. i., Cor. 1);
= ——· Similarly, b = and c =	.
P	P	p'" ’
/	»	\ ο	2Δ	2δ 2a	A δ	a
.-.(« +4 +ή, or 2.	=	—	+	— + _,	- + -	+
ΔΑΔΑ , ,
+ -77 + -FT,; aHd hence
r p p P
1=JL 1	_1_
r p' p"+pf"'
(2) (,-a) r’ = A (IV. IY„ Ex. 10); .·. («-«) = Again,
from (1) we haye (4 + e - a) = ~ +	^; but (4 + <· _ el
p V” p	r«οοκ τι.J	EXERCISES ON EUCLID.
211
12. Let ABC be a given Δ, and P a given point in one of the
aides. It is required to inscribe in ABC a Δ equiangular to DEF,
and having one of its angular points at P.
Sol.—From P let fall a _L PC on AB. Make the L PGH
= EDF, and GPH = DEF. Erect HI _L to PH, meeting BC in
I; join PI, and make the L JPI = GPH, and join IJ. JPI is
.the Δ required.
Dem. —Because the L GPH = JPI,. *. GP J = HPI, and the right
L PGJ = PHI; hence the Δ* PGJ, PHI are equiangular; .·. GP
: PJ : : HP : PI; alternation, GP : HP : : PJ : PI, and the
L GPH = JPI; hence (vi.) the Δ8 GPH, JPI are equiangular;
hut GPH, DEF are equiangular. Hence JPI is equiangular to
DEF, and it has one of its angles at the given point P.
13. Let ABC be a given Δ, and D, E, F three fixed points
its sides. It is required to prove that the locus of any point 0
its plane is a circle.
A
P
C
p 2
S’ 5*212
EXERCISES ON EUCLID.
[book vi-
Dem.—Join OB, OC, DF, DE. Describe Θ» about the X··
DBF, DCE, cutting OB, OC in G, H. Join GH. Npw since?
the points D, F are given, the line DF is given, and the L DBF
is given (hyp.); hence the Θ about DBF is given, and the-
L DBO is given by the given conditions; hence the arc DG
is given, and therefore G is a given point. In like manner H is
a given point; the line GH is given, and the L GOH is
given. Hence the locus of 0 is a circle.
14. (1) Dem.—Let the point B move along DE. From A let
fall a X AD on DE. Draw AF, making the L DAF = CAB~
From C draw CF X to AC, and let fall a X CG on AB. Now
because the L CAG is given, and the L AGC is a right L, the*
A ACG is given in species; therefore the ratio AC: CG is given ~rJtOOK VI.]
EXERCISES ON EUCLID.
213
Renee the ratio AC. AB : CGr. AB is given; but CO. AB is given,
therefore AC. AB is given.
Again, since the L DAF = BAC, DAB = CAF, and the1
Tight L ADB = ACF; therefore the Δ" DAB, CAF are equi-
angular; hence AD : AB :: AC : AF, .·. AB. AC = AD. AF; but
AB.AC is given, .·. AD.AF is given, and AD is given, .*. AF
-is given; and since the L DAF is given, .·. AF is given in posi-
tion, and the L ACF is right. Hence the locus of C is a circle.
(2) Let the point B move along a Θ. Produce AB to meet the
«circumference in D. Let 0 be the centre. Join OA, OD. Make
the L EAO = CAB, and ACE = ADO
Now (1) the rectangle AB . AC is given, and AB . AD is
given (III. xxxvi.); therefore the ratio AB.AC : AB.AD is
given; the ratio AC : AD is given. Again, since the Δ·
ACE, ADO are equiangular ; .·. AC : AE :: AD : AO; alter-
nation, AC : AD : : AE : AO; but the ratio AC : AD is given,
C
the ratio AE : AO is given; and AO is given, since it is
drawn from a fixed point to the centre of a fixed Θ ; .*. AE is
given in magnitude, and it is given in position, because it is
drawn making a given L with a given line; hence the point E
is given. And because the Δβ AOD, AEC are equiangular,
AO : OD : : AE : EC ; but the ratio AO : OD is given; .*. the
ratio AE : EC is given, and AE is given; EC is given, and
the point E has been shown to be given. Hence the locus of
O is a Θ, having E as centre and EC as radius.
15. (t) Let the vertex A remain fixed. Let the locus of B be
a right line DB. It is required to find the locus of C.
Sol.—From A let fall a A AD on DB. Make the / DAG
= CAB. Let fall CG ± on AG, and join DG.214	EXERCISES ON EUCLID. [book yi.
Now because the L CAB = DAG, the L C AG=DAB, and the right
Z CGA = BDA; hence the Δ* CAG, DAB are equiangular; AC
: AG : : AB : AD; alternation, AC : AB : : AG : AD; but the-
ratio AC : AB is given, since the Δ ABC is given in species; .·. the
ratio AG : AD is given, and AD is given in magnitude, because
it is a _L from a given point on a given line; .·. AG is given in
magnitude, and it is also given in position, since the Z DAG is
equal to a given Z CAB ; .·. G is a fixed point, and CG is at
right Z* to a given line at a given point. Hence the locus of C
is the line CG.
(2) Let the point B move along a Θ ; let 0 be its centre. Join
AO, BO, and draw AD, making the Z DAO = CAB. Draw CD,
making the Z ACD = ABO. Now the Δ ■ ACD, ABO are equi-
angular ; . *. AC : AD : : AB : AO; alternation, AO : AB : : AI>
: AO ; but the ratio AC : AB is given; .*. the ratio AD : AO is-
given, and AO is given; .·. AD is given. And since it makes
the Z DAO = CAB with a given line AO, .·. AD is given in posi-
tion ; hence the point D is given. Again, in the Δ * AOB, ADC
we have AO : OB : : AD : DC ·, but the ratio AO : OB is given;
the ratio AD : DC is given, and AD is given; .·. DC is given,
and the point D is given. Hence the locus of C is a Θ, having
D as centre and DC as radius.
16. (1) Bern.—Bisect the sides BC, CA, AB in D, E, F. Join
AD, BE, CF; let them intersect in 0. Produce AD to G, so that
DG = OD. Join BG. Draw EH || to AG, and produce BG to>
meet it in H.
Now since BD = CD, the Δ BDO = CDO, and the Δ BDA
= CDA; .*. the Δ BOA = COA. In like manner, COA = COBr
.*. the Δ* AOB, AOC, BOC are equal; .*. AOB = £ ABC. And
because OG = OA, the Δ BOG = AOB ; hence BOG = £ ABC.
And since the Δ* BOG, BEH are similar, BOG : ΒΈΗ : : OB*
: BE* (xix.); .*. BOG : BEH : : 4 : 9; that is, £ ABC : BEH
; : 4 : 9; hence 4 BEH = 3 ABC, .·. ABC = £ BEH. Again, it
is evident that the sides of the Δ BEH are equal to the medians
of ABC; hence, denoting the medians by a, J3, 7, and their half
sum by <r, we have (IV. iv., Ex. 12) the Δ BEH
«= VV. <r — a . σ — β . σ — y.
Hence the Δ ABC is equal to
£\/<r. <r — α.σ — £.<r — 7.BOOK VI.]
EXERCISES ON EUCLID.
215
(2) Dem.—Let Δ denote the area of the triangle; then (IV. rv.,
Ex. 4)Δ2=«.#-α.β-ί.β-<?; .·. Ι6Δ2= (a + b + c) (ft + e - a)
(c + a - b) (e+ b - <?).
Again, denoting the JLe by p\ p"> p”\ we have ap' = 2 Δ,
2 Δ 2 Δ 2 Δ
hp” — 2 Δ, and cp'" = 2 Δ; .·. («+£ + <?) = -y" + yT +
— 2 Δ + Ar +	; and, substituting, we get
\P P P /
16Δ3 = 2Δ|~+4+^]2Δί^ + ^-·~]
(p	P	P )	(p	P	Pi
I 1	1	1	)Λ	( 1	1	1 )
ip P	P )	(p	P	Pi
hence
and hence
Δ

17.	Let the Θ· ABC, DBE touch at B. Draw a common tan-
gent AD. Join, AB, DB, and produce them to meet the Θ8 in
E, C. Join DE, AC. DE, AC are the diameters of the Θ*
(III. xiii., Ex. 4).
Now the L ADC = AED (III. xxxii.), and the right L CAD
= ADE; therefore the Δ· CAD, ADE are equiangular. Hence
CA : AD : : AD : DE; that is, AD is a mean proportional be-
tween AC and DE.
18.	Let CL, OM, FN be the three parallel lines. Take any
point O in OM. Join AO, BO, and produce them to meet FN,
CL in D, E. Join AB, cutting the ||· in L, Μ, N. Join CF, and
produce it to meet AB produced in G. It is required to show that
G is a given point.
Now in the Δ AOB the line CFG cuts the three sides in
C, F, ’G; hence (“Sequel,” Book VI., Prop, iv., Sect, i.),
AC OF BG
CO * FB * GA
= 1; but ^	(n.), and the ratio
LU DM
AL
LM216
EXERCISES ON EUCLID.
[book VI.
AC .	.	T ..«	OF .	.	BO
given;	is given. In like manner, — is given; .·. —^
is given. Hence the line AB is divided externally in G in a
given ratio; . ·. G is a given point. Hence CF passes through a
fixed point. Similarly, DE passes through a fixed point.
19. Let a system of Θβ pass through two fixed points A, B.
From A draw any two secants, cutting the Θ* in C, D, E;
C', D', E'. It is required to prove that CD : DE : : C'D': D'E'.
Dem.—Join BC, BD, BE ; BC', BD', BE'.
Now the L ACB = AC'B (III. xxi.); .·. DCB ~ D'C'B, and
CDB = C'D'B; the Δ· CDB, C'D'B are equiangular; hence
CD : DB : : C'D' : D'B. In like manner, the Δ8 DEB, D'E'B
are equiangular, and BD : DE : : BD' : D'E'. Hence ex aequali
CD : DE : : C'D' : D'E'.
20. Let ABC he a Δ, the sides being denoted by a9 b, e. It is
required to find a point 0 in ABC, such that the diameters of theSOOK VI.]
EXERCISES ON EUCLID.
217
O* about the Δ8 OAB, OBC, OCA may be in the ratios of three
given lines l, m, n.
SoL—Construct a Δ EDF whose sides EF, DF, DE shall be
in the ratios	Produce ED to I, J. On CB describe a
l m n
segment of a Θ containing an L = IDF, and on AC a segment con-
taining anZ= JEF. 0, where these segments intersect, is the
required point.
Dem.—Join OA, OB, OC. Produce AO, and draw BG j| to
OC. From 0 let fall a A OH on BG. Draw CB, CT, the
diameters of the Θ8. Join BR, AT.
Now the sum of the L · AOC, GOC is two right L *, and the
sum of FEJ, FED is two right L 8; hence GOC = FED; but
GOC = OGB (I. xxix.); .·. OGB = FED. Again, the Z8 COB,
GBO equal two right Z 8, and IDF, EDF equal two right Z8;
GBO = EDF. Hence the Δ8 OBG, DEF are equiangular.
Because the L8 CTA, CO A = two right Z8 (III. xxn.),
^nd COA, COG equal two right Z.s, the L COG = CTA;
.·. OGH = CTA, and OHG = CAT, each being right; the
CT OG
Δ8 CAT, OGH are equiangular;	^	Again, the218
EXEECISES ON EUCLID.
[book VI.-
L8 COB, OBG equal two right L ·, and COB, CRB equal two
right Ls; .·. OBH = CRB, and the right L CBR == OHB;
.*. the Δ® CBR, OHB are equiangular;
CT CR
b * a
OG : OB; hut OG : OB
CR OB
* CB " OH
a b
I'm*
Hence
CT CR
b ' a
λ b CT CR
Hence CR : CT :: l: m. In like man-
l m ml
ner it can be shown that CT is to the diameter of the Θ about
OAB as m to n.
21.	Sol.—Describe a Θ about ABCD. Join CB, CD, BD.
Divide BD at E in a given ratio, and join CE, AC.
Now the points A, C are given, .·. AC is given in position, and
AD is given in position; hence the L DAC is given; hut (III. xxi.)
DAC = DBC; .·. DBC is a given L. In like manner, the L BDC
is given, .·. the L DCB is given; hence the Δ DBC is given in
species; DB : BC is given, and DB : BE is given (hyp.);
.·. BC : BE is given, and the L CBE is given. Hence the
Δ EBC is given in species. Now EBC is a Δ of given form.
One of its vertices, C, is fixed; another, B, moves along a line AB.
Hence (Ex. 15) the locus of E is a straight line.
22.	Dem.—Draw DF |j to EB. Produce CF, AD to meet
in H.
Now, because DF is || to BG, we have DF : BG ;: CF ; CB ;
hut DF = BF ; .·. BF : BG :: CF : CB.
C
Again, since the lines AC, BE, FD are parallel, we have
(ii., Ex. 1) BF : DE :: CF : AD; and, by similar Δ8, DE : EG
:: AD : AC; hence, ex aequali, BF : EG :: CF : AC ; but ACBOOK ΤΙ.]
EXERCISES ON EUCLID.
219*
= CB; BF : EG :: CF : CB. But it lias been proved that
BF : BG :: CF : CB, therefore BG = EG.
Lemma.—Take any point 0 within a Δ ABC. Join OA, OB,.
OC, and produce AO to meet BC in A'. It is required to prove
that the Δ OBC : ABC :: OA' : AA'·
Bern.—From A, 0 let fall ±s AD, OE on BC.
Now the Δ ABC = JBC.AD, and the Δ OBC = JBC.OE;.
hence ABC : OBC :: AD : OE; but AD : OE :: AA': OA;
.·. ABC : OBC :: AA': OA'.
23. Dem.—The Δ8 OBC + OCA + OAB = ABC. Divide by
ABC, and we have
OBC OCA OAB
ABC + ABC + ABC
.OBC OA'
^‘ab^aa1
and similarly for the others. Hence
OA' OB' οσ
AA' + BB' + CC “ *·
24. Dem.—AB : BC :: Δ AOB : BOC (i.), and A'B' : B'C'
:: Δ A'OB : Δ BOC'; .·. (Book V., Ex. 5)
AB BC AOB BOC
A'B' : B C' 1; A'OB' ; BOC';
but (xxm., Ex. 1),
AOB BOC	AO.OB	OB.OC	AB	BC
ΑΌΒ' : B OC' :: A'O.OB' : OB'.OC'’ *** A'B' : B'C'
AO.OB	OB.OC	AB	BC	AO	OC
:: A'O.OB' :	OB'.OC,; * *	A'B'	: B'C' **** A'O ** OC'*
Hence
AB OC _ BC OA
A'B'*OC'"~B'C'*OA’
And similarly,
BC OA'	CA	OB
B'C'OA ~ C'A' * OB'*
25. (1) Dem.—Draw the diagonals AC, BD. Bisect them in
F, E. Join FE, and produce both ways to meet AD, BC, and
DC produced in H, G, I. Now, in the Δ BDC, the line ElEXERCISES ON EUCLID.
220
[book ΎΙ.
euts the three sides in E, G, I. Hence (“ Sequel,” Book YI.
Prop, it., Sect, i.)
BE DI CG_
ED'IC'GB ~ 1;
but
BE .	DI _ GB
ED ~~ ’ ·’’ IC*GB~1; *’· IC“ CG*
In like manner, from the Δ ADC, we get
DI_ HD
IC ~ AH*
Hence
GB _ HD
CG “ AH*
(2) Dem.—Join 0, the centre, to A, B, C, D. Join 0 to J, K,
A
"where AD, BC touch the Θ. Now, since OK = OJ, we hare
<i.) the A'OBC : OAD :: BC : AD. Let fall 1» BL, DM on
•OG, OH; then (I. xxvi.) the Δ· BEL, DEM are equal; BL
— DM, and .·. the Δ OBG : OHD : : OG : OH. In like man-
ner OCG : OHA :: OG : OH. Adding, we have OBC : OAD
:: OG : OH; hut it was shown that OBC : OAD :: BC : AD.
Hence BC : AD :: OG : OH.
(3) Dem.—Consider the Δ ECI. It is intersected by AB;
hence (“Sequel,” Book YI., Prop, iv., Sect, i.)
EG IB CA	CA	EG IB
GI *BC*AE ; butAE ; *** GI ’ BC **
Again, consider the Δ AEK; it is intersected by CD ;
EH KD AC .EXERCISES ON EUCLID.
221
JBOOK TI.]
and, as before,
EH KD _ 1 EG IB _ EH KD
HK’DA""2’ ' * GrT’BC HK# DA*
Now AD, BC are opposite sides, and they are cut by EF in K, I p
hence (1) they are cut proportionally;
CB AD	EG _ EH
·’’ IB “DK’and ·’* GI “HK;
A
that is, EG : GI : : EH : HK; and the first is to the sum of the
first and second as the third is to the sum of the third and fourth.
Hence EG : ΈΙ: : EH : EE.
26. It is required to prove that AD . DB : AC . CB :: AD®
: AC2.
Dexn.—AD . DB, AC . CB, are rectangular figures; and since
AD : DB : : AC : CB (hi.), these figures are similar; hence
(xix.) AD. DB : AC. CB : : AD2 : AC2. In like manner AC. CB
: AD'. D'B :: AC2 : AD'2.
(1) Dem.—If AD . DB, AC. CB, and AD'. D'B, are in A. P.r
the difference between AD . DB and AC . CB is equal to the
difference between AC. CB and AD'. D'B; but AC. CB - AD. DB
= CD2 (xvn., Ex. 1), and AD' . D'B - AC . CB = CD'2; .·. CD2
= CD'2, . ·. CD = CD', .·. the L CDD' = CD'D; but the
L DCD' is right; .*. each of the L· CDD', CD'D is half a
right L; hence the L CDA is a right L and a-half. Now the
L CDA = CBD + BCD, and CDB = CAD + ACD ; hence CDA
— CDB = CBD - CAD i but the difference between CDA and
CDB is a right L. Hence the difference between CBD and
CAD is a right /.·222
EXERCISES ON EUCLID.
[book ti.
(2)	Dem.—If the three rectangles be in G. P., the squares of
the lines DB, BC, BD' are in G. P.; .·. DB, BC, BD' arq in G. P.,
.·. BC is a mean proportional between DB and BD'; hut the
_L is a mean proportional between the segments of the hypote-
nuse (tiii., Cor. 1). Hence BC is a X, and hence the L ABC
is right.
(3)	Dem.—If the rectangles AD.DB, AC.CB, AD'. D'B are
in Η. P., the 1st: 3rd :: difference between 1st and 2nd : differ-
ence between 2nd and 3rd; hut difference between 1st and 2nd
= CD2 (xvn., Ex. 1) and difference between 2nd and 3rd = CD'2,
.·. AD. DB : AD'. D'B :: CD2: CD'2; but, by similar figures,
AD.DB : AD'.D'B :: DB2 : D'B2; hence CD2 : CD'2 :: DB2
: D'B2; .*. CD : CD' :: DB : D'B, and .·. (iii.) the L DCD' is
bisected,. ·. the L DCB is half a right L; but the L ACD = DCB;
.·. the L ACB is right. Hence the sum of the L8 CAB, CBA is
a right L.
28.	Dem.—Denote the radii of the Θ8 by p, p ; then (VI. rr.)
DC : D'C :: p : p', and A'C : BC :: p : p', .*. DC : D'C :: A'C
: BC; DD* : D'C : : A'B : BC. (V. xvii.) In like manner
DD' : D'C :: AB' : B'C, .·. DD'2 : D'C2 :: A'B.AB' : BC.B'C;
but D'C2 = BC. B'C (III. xxxvi.). Hence DD'2 = A'B.AB'.
29.	Dem.—Because A'O is H to BO", AO" : 00" :: AB : A'B
(n.); that is, R : (R — p) :: AB : A'B. Similarly, R : (R — p')
:: AB : AB'; .·. R2 : (R - p) (R - p') :: AB2 : A'B.AB'; but
A'B.AB' = DD'2 (III. xxxvi.). Hence R2 : (R - p) (R - p')
:: AB2 : DD'2.
30.	Dem.—Let A, B, C, D be the points in which the four Θ8
touch the fifth. Join AB, BC, CD, DA, AC, BD, and denote
the radii of the four Θ* by pi, p2, pz* pu and the radius of the
fifth by R; then putting 122 for DD'2, we have, from Ex. 29,
AB2 : 122 :: R2 : (R — pi) (R - p2); hence
12. R
V(E-P1) (E -
Similarly,
34. R	t ^	14 . R
CD
V (R - pi) (R - p4)BOOK VI.]
EXERCISES ON EUCLID.
223
Έ	23. R
and	BC = ---------
Now, by Ptolemy’s theorem (xvn., Ex. 13) AB.CD + BC.AD
= AC. BD. Therefore
_________12.31. R*________ 23.14. R*
V^R-Pl)(R-p2)(R-pa)(R-p4) + V(R-ps)(R -P8)(R -p,)(R-P4)
____________Ϊ3.24. Ra_________
" V(R-p,)(R-p3)(R-p2)(R_pi):
and hence
12.34 + 23.14= 13.24.
31. Dem.—Bisect the sides of the Δ ABC in the points D, E,
F. Inscribe a Θ in ABC, touching the sides in G, Η, I. Let
the sides opposite the angular points be denoted by a, b, c.
Now if we consider the points D, E, F as infinitely small ©*,
DE, EF, FG are common tangents to the Θ8 1, 2 ; 2, 3 ; 3, 1;
hence we have 12 = DE = J AB = Similarly, 23 = ^ a,
31 = % b.
Let the inscribed Θ be denoted by 4. Now BD = £ BC
= i a, and BG = (s — b) (IV. iv., Ex. 2); .·. DG - i a — (s — b)
= i(b — J) ; that is, 14 = J (b - c). In like manner, 24 = \(e - a),
and 34 =	— 5). Now if we substitute these values in the con-
ditions of the last question, we find that it is fulfilled. Hence the224
EXERCISES ON EUCLID.
[book ti.
Θ through the middle points of the sides of the Δ touches the
inscribed circle. Similarly, it touches the escribed circle.
32. Let A, B, C, D be the four points; join them, and join
AC, BD. Bisect BC, BD, CD in E, F, G. Bisect AB, AD in
Η, I. Describe a Θ through the points E, F, G, and another
O through Η, I, F; let them intersect in J. It is required to
prove that the Θ8 through the middle points of the Δ8 ABC,
ADC will also pass through J.
Dem.—Bisect AC in K. Join KE, KH, EH, GE, EJ, JF,
H, HI, IF, JK.
Now because CB, CD are bisected in E, G, EG is || toJBD.
Similarly, GF is || to BC ; hence BEGF is a parallelogram -r
the L FGE = FBE ; but FGE = FJE (III. xxi.),
FJE ss FBE. Again, as before, HIFB is a parallelo-
gram, .·. the L HIF = HBF; but HIF = HJF (III. xxi.HHOOK VI.]
EXERCISES ON EUCLID.
HJF = HBF; .·. the whole L HJE = HBE; but HBE
= HKE, since HKEB is evidently a parallelogram; .*. HJE
= HKE; hence the four points Η, K, J, E are concyclic, and
the Θ through Η, K, E will pass through J. Similarly, the 0
through K, I, G will pass through J. Hence the four nine-points
0s have a common point.
33. Dem.—From A let fall _L8 AE, AF, AG on CD, PJB, CB.
Now because the L8 AED, AFD are right, AEDF is a cyclic
quadrilateral, and AD is the diameter of its circumcircler Draw
another diameter EH. Join EF, FH. About the Δ BDC
describe a ©. Draw its diameter BI, and join IC. Now
(III. xxii.) the sum of theZ.8 EHF, EDF is two rightL*>
and the sum of EDB, CDB is two right L ■; hence the L EHF
= CDB; but (III. xxi.) CDB = CIB; hence EHF = CIB, and
the right L EFH = ICB; .·. the Δ8 EFH, ICB are equiangular;
hence EH : EF : : IB : BC ; EH. BC = EF. IB ; that is, at
— EF.IB. Similarly, bd = FG.IB, and DD' = EG.IB. Hence
EF, FG, EG are proportional to ac, bd, DD'.
34. OEDF is a four-sided figure; OD, EF its diagonals. If
OF.DE + OE.DF = OD.EF, it is required to prove that
OEDF is a cyclic quadrilateral.
Dem.—Produce OD, OE, OF to B, C, A until each of the
rectangles OD. OB, OE. OC, OF . OA is equal to the square of
a given lint, say R2. Join AB, BC, AC.
Now’OD .OB = OE.OC; .·. OB : OC : : OE : OD, and the
L ROC is common to the two Δ8 OBC, OED; hence (vi.) they
are equiangular, and BC: OB:: ED: OE; alternation, BC : ED::
Q226
EXERCISES ON EUCLID.
[book n.
OB: OE; .·. BC : ED :: OB.OD: OE. OD; that is, BC: ED::
ED BC T	DF
oe7od= R8 ’ In llke manner-
AB , EF AC
R2 : OE. OD; hence -
OF.OD
= “^oITTOF = Now(hyp.)ED.OF+DF.OE=OD.
ED DF _ EF Λί. BO , AB
OE.OD + OD . OF OE.OF ’ 0144 W +W~
EF;
AC
R*
; .·. BC + AB = AC ; but this could not be true unless AB
O
and BC are in one straight line; .·. ABC is a straight line; .·.
the sum of the L* ABO, CBO is two right L*; hut ABO = DFO,
and CBO = DEO ; .·. DFO + DEO = to two right L%· And
hence OEDF is a cyclic quadrilateral.
Ltmma.—If C be the external centre of similitude of two Θ';
CH any line passing through C, and cutting both Θ' in the points
E, F; G, H; it is required to prove that CG. FC = AC . BC.
Bern—Join AG, DE.
Now AC : DO :: GC : EC; .·. AC.BC : BC.DC :: GC.FC:
FC.FC ; but BC.DC = EC.FC. Hence AC.BC = GC.FC.
35. (1) Let 0,0' be the centres of the given Θ», and P the point.HOOK τι.]
EXERCISES ON EUCLID.
227
Sol.—Join 00', and produce. Let C be the external centre of
similitude. Join PC, and find the point Q, so that PC. QC
= AC . BC. Describe a Θ passing through P, Q, and touching
Hi Θ whose centre is O in G (III. xxxvn., Ex. 1). This is
the required circle.
Dem.—Join GC, cutting the circle whose centre is 0' in F.
Now (const.) PC. QC = AC.BC, and(Zmiiw) AC.jBC = GC.
FC; .·. PC.QC = GC.FC. Hence the Θ through the points
P, Q> G passes through F, and touches the Θ whosejcentre is O'.
(2) Sol.—Let 0, O', 0" be the centres of the given Θ8. Draw
any two radii OA, O'B. Cut off AC, BD, each equal to the radius
of 0". With 0 as centre and OC as radius, describe a Θ. With
O' as centre and O'D as radius, describe a Θ. Now (1) describe
a Θ touching those two in E, F, and passing through the point 0".
Let 0"'be its centre. Join 0'"0,0"'0', 0'"0", and produce them
to meet the circumference of the given Θ8 in the points G, Η, I.
The Θ through G, Η, I will be the required circle.
Q 2228
EXERCISES ON EUCLID.
[book ti.
Dem.—Because 00 = OA and OE = OC, EG = AC; but AC
=: 0"I; EG = Ο'Ί, and 0"'E = 0'"0" ; hence 0"'G = 0"'I.
In like maner, 0"'H = 0"'I. Hence the Θ described with O'"
as centre, and 0"'G as radius, will pass through Η, I, and touch
the given Θ8 in the points G, Η, I.
36.	Let 0, O' be the centres of the fixed Θ8, and C their centra
of similitude; and let any variable Θ 0" touch 0, O' in G, F.
From C draw CD a tangent to 0". It is required to prove that
CD is of constant length. (See Diagram to Ex. 35 (1)).
Dem.—Join GF, and produce it to pass through C.
Now CD2 = GC.CF (III. xxxvi.), and GC.CF = AC.CB
(Lemma to 35); hence CD2 = AC. CB ; but AC. CB is constant,,
since A, C, B are fixed points. Hence CD is constant.
37.	Dem.—Draw DD' a common tangent to the two fixed Θ8.
Join AD, BD', and produce them; they must meet on the cir-
cumference of 0". For, if not, let AD meet the circumference
of 0" in P, and BD' meet it in Q. Join 0"0, 0"0', and produce
them; O'O, 0"0' must pass through A, B (III. xi.). Join OD,.
O'D', 0"P, 0"Q. Now the L 0"AP = 0"PA, andOAD = ODA;
.·. ODA = 0"PA; hence OD is parallel to 0"P. Now the
L ODD' is right (III. xviii.) ; hence 0"P is X to DD'. Simi-
larly, 0"Q is X to DD', which is impossible, unless Q coincide
with F. Hence BD' must pass through P.
38.	Join A'B'. Take a fixed point C in AC, and in BD find a
point D, so that as AA': AC :: BB': BD. Join AB, and divide
it in E in a given ratio. Join CD, and divide it in F in the same
ratio. Join EF, cutting A'B' in 0. It is required to prove that
A'O : OB':: AE : EB.UOOK VI.]
EXERCISES ON EUCLID.
229
Dem.—Through F draw GH parallel to AB, and draw AG,
BH, each parallel to EF. Join CG, DH. Draw AT parallel to
AG, and B'J parallel to BH. Join IF, JF.
Now, by construction, AA': AC :: BB' : BD ; .·. AC : A'C
^ : BD : B'D. And hence, by similar triangles, GC : IC :: DH:
DJ; but GC : CF :: DH : DF. Hence IC : CF :: DJ : DF,
and the contained angles ICF, JDF are equal, .·. the triangles
ICF, JDF are equiangular, .*. the L IFC = JFD; .*. IF, FJ
are in the same straight line.
Again, from similar A8, AG : AT : : AC : A'C, and BH : B'J
:: DB': DB; hence AG : AT :: BH : B'J; but AG = BH ;
A'l = B'J; hence IJ is parallel to A'B'; .*. AO': OB' :: IF :
FJ; that is, :: CF : FD, or :: AE : EB. Hence the locus of
the point in which A'B' is divided in the ratio of AE : EB is the
right line EF.
39.	Dem.—It was proved in the last Exercise that A'O : OB
^ : AE : EB. In like manner, EO : OF :: AA' : A'C. Now
putting G, H for A', B', we have GO : OH :: AE : EB, and
EO : OF :: AG : GC.
Lemma.—If a given line AC be divided in B, so that AB.BC4
is a maximum; it is required to prove that BC = 4 AB.
Dem.—Divide BC into four equal parts in E, F, G; then
BC
each of the parts BE, EF, FG, GC is equal to —; hence
4*
BC4
BE.EF.FG.GC =	Multiply each by AB, and we get
ΑΒΕΓ G C
AB.BE.EF.FG.GC =^221; but (hyp.) AB. BC* is a maxi-
mum; . ·. AB. BE. EF. FG. GC is a maximum; .·. AB, BE, EF,
FG, GC are all equal (“ Sequel,” Book II., Prop, xii., Cor.).
Hence BC = 4 AB.
Similarly, if it be required to divide AC in B, so that AB. BO
may be a maximum, BC = hAB.
40.	Analysis.—Let ABC be the required Δ. Bisect the vertical
L ABC by BH. From A, C let fall _L8 AD, CF on BH, and
from B let fall a 1 BE on AC. Join DE, EF. Draw HI, the
diameter. Join BI. Draw BK || to AC, and let fall a ± EG
on HB.230
EXERCISES ON EUCLID.
[book VI-
Now the L ADB = AEB, each being right; hence the four
points A, D, E, B are concyclic; .*. the L EDF = BAC. Again,
because each of the L8 BEC, BFC is right, BFEC is a cyclic
quadrilateral; .·. the sum of the L9 BFE, BCE is two right L8,
and the sum of BFE, DFE is two right L8; .·. the L DFE
= BCA; .*. the Δ8 ABC, DEF are equiangular. And since
their _L8 are BE, EG, ABC : DEF : : BE2 : EG2; but BE2 r
EG2 : : HI2 : IB2, or HI. IK ; .·. BE2 : EG2 : : HI : IK;
.·. ABC : DEF : : HI: IK; .♦. ABC . IK = DEF. HI. Now
DEF is a maximum (hyp.), and HI is a given line, because
it. is the diameter of the Θ; .*. ABC. IK is a maximum.
Now ABC = J base. perpendicular = AL. BE, or AL.KL;
.·. AL. KL . IK is a maximum. Now whatever AL is, the rect-
angle KL.IK is a maximum when IL is bisected in K, and then
KL.KI = £IL2;
at IL2 ·
·. AL. is a maximum ;
4
AL. IL2 is a
maximum; .*. AL2 . IL4 is a maximum; but AL2 = HL . LI;
.·. HL.IL5 is a maximum. And .*. {Lemma) IL = 5 HL.
Hence the method of construction is evident.
41. Let AC, BD, the diagonals of the inscribed quadrilateral,,
intersect in 0. At the points A, B, C, D draw tangents to tho
Θ. Let them meet in E, F, G, H; then EFGH is a circum-
scribed quadrilateral. It is required to prove that its diagonals
EG, FH must pass through 0.
Dem.—If possible let EG not pass through 0 ; but cut AC,
BD in I, K. Produce AE, CF to meet in J (not represented in
the diagram). Through E draw EL || to GJ, and EM || to GD.BOOK VI.]
EXERCISES ON EUCLID.
231
Produce DB to meet EM. Now because JA= JC being tangents,
the£ JCA = JAC ; butELA = JCA(I. xxix.); .·. EAL = ELA ;
and EA = EL. In like manner EB = EM; but EA = EB ;
.·. EL = EM. Now since the Δ8 GCI, ELI are equiangular,
GC : GI:: EL : El; alternation, GC : EL :: GI: El; but GC =
GD, and EL = EM; .·. GD : EM : : GI: El; and because the
Δ· GKD, MKE are equiangular, GD : EM:: GK: EK; GI :
El : : GK : EK, which is impossible unless the points I, K
coincide. Hence GE must pass through 0. In like manner FH
must pass through 0.232
EXEBCISES ON EUCLID.
[book VI.
and X the given line. Through A, B describe any Θ cutting
W in C, D. Join AB, CD, and produce them to meet in E.
Through E draw EFG parallel to X, and cutting W in F, G.
The Θ through A, B, F, G is the one required.
Dem.—AE.EB=CE. ED,andCE.ED=GE. EF; .·. AE.EB
= GE. EF. Hence the four points A, B, F, G are concyclic, and
the common chord FG is || to X.
(2) Sol.—Let 0 he the given point. Make the same construc-
tion as before; and instead of drawing EFG || to X, join EO,
and produce it to cut Win F, G. Then, as in (1), EFG is a
common chord, and it passes through 0, the given point.
43. Sol.—Let 0 be the centre of the Θ, ABC the Z, and DE
the given line. Produce AB, CB to meet DE in E, G. Bisect
GE in F. Join FB. From 0 let fall a _L OD on DE, and
meeting FB produced in H. Through H draw IJ || to DE,
A
meeting AB, CB in I, J. Join 01, OJ. Now because the lines
GJ, FH, El pass through B, and are cut by the ||· GE, IJ,
GF : FE : : IH : HJ; but GF = FE ; IH = HJ ; and since
IJ is || to DE, and OD meets them, the Z OHJ = ODE; .*. OHJ
is a right L ; ·'· OHI is right, and .*. (I. rv.) OJ = 01; and the
Θ, with 0 as centre, and OJ as radius, will pass through I, and
its chord IJ is parallel to the given line DE.
44. Let ABCDE be a polygon of an odd number of sides.
Take any point 0 within it. Join AO,|BO, CO, DO, EO, and
produce them to meet the opposite sides in A', B', C', D', E'. It
is required to prove that the product of AD', BE', CA', DB', EC'
is equal to the product of A'D, ΒΈ, C'A, D'B, E'C.
Dem.—Join AC, AD. Now the Δ AOD : A'OD :; AO: ΑΌ
(i.), and AOC : AOC : : AO : ΑΌ; .·. AOD ; A'OD AOCBOOK VI.]
EXERCISES ON EUCLID.
23S
: A'OC ; alternation, AOD : AOC :: AOD : A'OC; but AOD:
A'OC :: A'D : A'C. Hence
A'D AOD
A'C “AOC*
In like manner, by joining BE, BD ; CE, CA; DB, DA; EC,
EB, we get
EF _ EOB # AC/ _ AOC e BD' BOD CE' _ COE
BD ~ BOD5 C'E “ EOC ; D'A “ DOA ; E'B " BOE*
Now, multiplying these together, we find that the numerators of
the second terms are equal to the denominators. Hence the pro-
duct of the numerators of the first terms is equal to the product of
the denominators ; that is, A'D. B'E. C'A. D'B. E'C = A'C. B'D.
C'E.D'A.E'B.
45.	Let ABC he the A, and let the sides touch the Θ in the
points Af, B', C'.
Dem.—Join A A', BB', CC'. Now AB' = AC', BA' = BC%
CA' = CB'. Hence AC'. CB'. BA'= A'C . C'B . B'A; and hence
(Ex. 4) the lines AA', BB', CC' are concurrent.
46.	From A, B, C draw tangents AA', BB', CC'; and produoeEXERCISES ON’ETJCLID.
234
[book. VI.
the sides BC, AC, AB to meet them in A', B', C\ It is required
to show that the points A', B', C' are collinear.
Dem.—The L B'BC = BAB' (III. xxxii.), and the L BB'C is
common, .·. the Z.8 AB'B, BB'C are equiangular, .*. AB' : AB
: : BB' : BC; alternation, AB' : BB' : : AB : BC, .*. AB'2 : BB'*
: : AB2 : BC2; hut BB'2 = AB'. B'C (III. xxxvi.), AB'2: AB.
B'C : : AB2 : BC2, .·. AB' : B'C : : AB2: BC2. Hence, denoting
the sides of the Δ ABC by λ, b, c, we have AB': B'C : : e2: a2.
Interchange, and we get BC' : C'A :: a2 : 2, and CA' : A'B
: zb2: c2. Multiply these together, and we have AB'. BC'. CA':
B'C.C'A.A'B : : &a2b2 z a2Pc2; .·. AB'.BC'.CA' = B'C.
C'A. A'B; and hence (Ex. 5) the points A', B', C' are col-
linear.
47. Dem.—Produce the sides, and draw AA', BB', CC', bisect-
ing the external L ·. Now (hi., Ex. 1 AB': B'C:: AB: BC. Inter-
change, and we have BC' : C'A :: BC : CA. Interchange again,
and CA' : A'B : : CA : AB. Now, multiply together, and
AB'. BC'. CA': B'C . C'A. A'B:: AB. BC . CA : BC . CA. AB ;
but the third term is equal to the fourth, .*. the first is equal to
tho second; that is, AB'. BC'. CA' = B'C . C'A. A'B ; and hence
(Ex. 5) the p oints A', B', C' are collinear.
Lemma.—Let two Θ8, whose centres are 0, O', cut in P. Join
OP, O'P. Produce O'P to E. Draw CP, DP tangents to the Θ*.
It is required to show that the L EPO = CPD.
Dem.—Produce CP, DP to F and G. Now the L O'PF
right (III. xvni.); hence (I. xv.) CPE is right, and OPD
right; .·. CPE = OPD. Reject OPC, and EPO = CPD.
£· S’KOOK τι.]
EXEBCISES ON EUCLID.
235*
48.	Let AB be a given line, P a given point, 0 the centre of
the given Θ, and X a given L. It is required to describe a Θ,
touching AB in P, and cutting 0 at an L equal to X.
Sol.—Erect PC _L to AB. Draw DP, making the L CPD =
Produce DP to E, cut off EP equal to the radius of 0. Join EO.
Bisect it in F. Erect FC _L to EO, meeting PC in C. With (J
as centre, and CP as radius, describe a Θ, cutting 0 in 0. This
is the required circle.
Bern.—Join EC, CO, CG, OG. Now because EF = OF, and
FC common, and the L EFC = OFC; .·. (I. iv.) EC = OC, and
CP = CG, being radii, and EP = OG (const.), .·. the L EPC
= OGC; but DPC and EPC are supplements ; and HGC, OGC
are supplements, .·. HGC = DPC ; but DPC = X, and HGC is
equal to the L between the Θ3 {Lemma). Hence the L between
the Osis equal to the given L, and the Θ PG touches AB in P.
49.	See “ Sequel,” Book IV. Prop, hi., Cor. 2.
50.	See “ Sequel,” Book I. Prop. xvn.
51.	See “ Sequel,” Book II. Prop. x.
52.	Let 0 be the centre of mean position of the feet of X8 from
it on the sides. From 0 let fall Xs OA', OB', OC' on the sides.
Take any other point P within the Δ, and let fall Xs PA", PB",
PC" It is required to show that OA'2 + 0B'2 + OC'2 is less than
PA"2 + PB"2 + PC"2.
Bern.—Join OP, PA', PB', PC'. Now, because 0 is the centre-
of mean position,of A', B', C', we have (Ex. 51) A'P2 -f B'P2236
EXERCISES ON EUCLID
[book \n.
4- C'P2 = OA'2 + OB'2 4 OC'2 + 3;OP2; but AT2 = A'A"2 + A"P»,
B'P2 = B'B"2 4 B'T2, and C'P2 = C'C"2 + C"P2; .·. A'A"2
4 B'B"2 4 C'C"2 4 A"P2 4 B"P2 4 C"P2 = OA'2 4 OB'2 + OC'»
4- 3 OP2.
From P let fall a X PD on OC', then OP2 is greater than PD2;
that is, greater than C'C"2. In like manner it is greater than
A'A"2, and greater than| B'B"2,	3 OP2 is greater than A'Arn
4 B'B"2 + C'C"2; and hence A"P2 + B"P2 + C"P2 is greater than
OA'2 + OB'2 + OC'2.
53. (1) Let A, B he the opposite Ls; w, n the diagonals, and
O the angle between the diagonals.
Sol.—Construct a parallelogram DEFG, haying two adjacent
aides DE, DG respectively equal to m and #», and their includedbook τι.]	EXERCISES ON EUCLID.	237
L « to C. On DE describe a segment of a Θ containing an
L equal to A; and on FG describe a segment containing an
L equal to B; let them intersect in H. Join HD, HE, HF,
HG. Through H draw HK || and = to EF. Join DK, EK.
DHEK is the required quadrilateral.
Dem.—The L DHE = A, and EF = HK (I. xxxiy.); but
EF = GD ; HK = GD, and it is || to it; HKDG is a.
parallelogram ; .·. HG is J| to DK, and HF is || to EK ;
hence the L GHF = DKE; but GHF = B, .*. DKE
= B; and (I. xxix.) the L HME = GDE; but GDE = C,.
HME = C.
54. Let a O, whose centre is O', roll inside another Θ, 0, whose
diameter is twice that of O'. Take a fixed point P in the circum-
ference of O'. It is required to find its locus.
Sol.—Let R be the point of contact. Join OP, OR, O'P, and
produce OP to meet the circumference in Q, and bisect the-
L RO'P by 0'S.
Now the L ROT = 2 ROP (III. xx.); .·. the L RO'S = ROQ,
and the arc RS : RQ: : O'R : OR; but OR = 2 O'R, .·. RQ = 2RS,
.·. RP = RQ. Now, since the arc RP = RQ, the point P must
have coincided withJQ. Hence the line OQ is the locus of P.
55. Sol.—Take any point G in the arc CD. Join CG, DG.
From the centre 0 let fall a JL OE on CD, and on OE describe a
segment OFE containing an L equal to CGD. Join OC. Bisect
itinH. Through H draw HF || to AB, cutting the segment OFE
in F. Join OF, and through C draw CP || to OF. P is the re-
quired point.238
EXERCISES ON EUCLID.
[book ti.
Dem.—Let CP intersect AB in K. Join PD, cutting AB in
J. Produce EH to meet CP in I. Join EE, and produce it to
meet CP in L. JoinCF, FJ. The points C, F, J are collinear;
if not, let CF, FM be in a straight line. Now (III. xxn.) the
L8 CGD, CPD equal two right L8; .·. OFE, CPD equal two right
L s, and OFE, OFL equal two right L 8, .·. OFL = CPD; that
is, CLE = CPD ; hence EL is || to PD. Again, in the Δ COM,
since CD is bisected in H, CM is bisected in (I. xl., Ex. 3) ; and
similarly, in the Δ CDN, CN is bisected in F; FN = FM,
which is absurd ; hence CF produced must pass through J, and
CF = FJ. Now, in the Δ CJK, CJ is bisected in F, and OF
is parallel to CP, .·. KJ is bisected in 0; that is, OK = OJ.
56. Let ABCD be a polygon of four sides. Produce AB, CD
to L, N, and draw a transversal LMON, cutting the four sides.
From A, B, C, D let fall 18 p\ p"t p'", p'"' on LMON.
CBOOK VI.]
EXERCISES ON EUCLID.
23£
Now, since the Δ8 Aj/L, Bp"L are equiangular,
	AL	P' /
	1^ II iw	
For the same reason,		
BM pn	' CN	p·"
CM“/'	7’ DN“	p" ’
Multiplying together,	we get	
l™.
AO
AL . BM. CN . DO pp"p'”p""
BL. CM. DM. AO = ρ>"ρΥ"#
Hence AL. BM . CN. DO = BL . CM. DN . AO. And similarly
for a figure of any number of sides.
57. Let the transversal LMN cut the sides of the Δ ABC in
the points L, Μ, N. Bisect LN, NM, ML in Ο, P, Q. Join AP,
OB, CQ, and produce them to meet the sides of the Δ ABC in
A', B', C', respectively. It is required to prove that the points
A', B', C' are collinear.
Dem.—The sides of the Δ AMN are cut by OB ; therefore
AB' MO NB
B'M' ON ’ BA —	1 (E 5)‘
1 And since the Δ CLM is <Sut by 0B',^£ .52	= _ i.
B C BL OM
• it u: i -	«	*1.	1. AB' CB NB .
Multiplying together, we have	gj-= 1; interchange,240
EXERCISES ON EUCLID.
[book vt.
.BC' AC LC , . A ,	. .OA'BAMA ,
“dC;A-CB-CM = 1; mterchangeagam,and —. — . — = 1.
Multiply these results together, and we get
AB' BC' CA' NB LC MA _
B/C‘C'A'A'B,BL,CM,AN"1;
NB LC MA _ Ί ■_ ,x..AB;BO; ca
BL ‘ CM ’ AN “	1	: '*· B'C ’ C'A" A*B —
And hence the joints A', B', C' are collinear.
58. Let ABC he the Δ. Join PA, PB, PC, and erect at P _L·
ΑΈ ΒΉ, C'G to PA, PB, PC, intersecting the sides BC, CA,
AB, respectively in A', B', C\ It is required to show that the
points A', B', C' are collinear.
Dem.—From A, B, C let fall ±* AG, AI on CG, ΒΉ; BDr
BF on ΑΈ, C'G; CE, CH on ΑΈ, B'H.
Now, because each of the L* APA% BPB' is right, the
L API = BPD, and AIP = BDP; hence tho As AIP, BDP
are equiangular. In Jike jnanner, the A* AGP, CEP are equi-
angular, and CPH, BPF are equiangular.book τι.]
EXERCISES ON EUCLID.
241
Again, since the Δ8 CA'E, BA'D are equiangular,
CA'
A'B
Similarly,
AB' _ AI . BC' BF
B'C “ CH* C'A ~ AG;
CE
BD*
therefore
CA'.AB'.BC' CE.AI.BF
A'B.B'C . C'A “ BD. CH. AG’
hence
CA'. AB'. BC'	CE. AI. BF. PB. PC. PA
A'B. B'C . C'A “ BD.CH. AG. PB. PC. PAJ
but AI. BP = BD. AP, since the Δ8 AIP, BDP are equiangular,
and PA. CE = AG.PC; and PC.BF = PB.CH; .·. CA'.AB'
. BC' = A'B. B'C. C'A. And hence (Ex. 4) the points A', B', C'
are collinear.
59. Let ACB he a given semicircle. It is required to divide it
into two parts by a JL on the diameter AB, so that the radii of
the Θ8 inscribed in them may have a given ratio DE : EF.
Sol.—On DF describe a segment containing an L equal to half
a right L. Erect EG 1 to DF. Join DG, FG. At the point F
in FG draw FH, making the L HFG = HGF. In the semicircle
B242	EXERCISES ON EUCLID. [book vi.
draw BC, making the L ABC = EFH. Let fall the 1 Cl on AB.
01 is the required line.
Dem.—In the figures CIBM, CIAN describe Θβ, touching
IB, IA, IC, and the arcs BC, AC in the points J, K, L, Μ, N.
Let 0, 0' be their centres. Join OJ, CJ, OL, O'L, AL, LM.
The points A, L, M are collinear (hi., Ex. 51). Join BM,
AC, CK. Now the L LIB is right, and LMB is right
(III. xxxi.); .*. ILMB is a cyclic quadrilateral; .*. BA . AI
= MA. AL; but BA. AI = AC2 (I. xlyii., Ex. 1), and MA. AL
= AJ2 (III. xxxyi.) ; .·. AC2 = AJ2; .·. AC = AJ; .*.the/.ACJ
= AJC ; .·. ACJ = JBC+JCB; but ACI = IBC (viii.) ; .·. ICJ
= BCJ. In like manner, the L ICK = ACK ; hence the L KCJ
is half a right L . Now in the A8 EHF, ICB the L BIC = FEH,
and IBC = EFH (const.); .·. ICB = EHF ; hut ICB = 2 ICJ,
and EHF = 2 EGF ; .·. ICJ = EGF, and CIJ = GEF; CJI
= GFE; hence the A8 CIJ, GEF are equiangular. And because
the L DGF = KCJ, and GFD = CJK; .·. GDF = CKJ; hence
the A8 CJK, GFD are equiangular; .·. ΚΙ : IJ :: DE : EF;
but KI : IJ : : O'L : OL. Hence O'L : OL :: DE : EF.
Lemma.—If A, B, C he fixed points, and P a variable point,
find the locus of P, if mAP2 + wBP2 + jt?CP2 is given.
Sol.—Join AP, BP, CP, AB, BC. Divide AB in D, so that
»«AD = wDB. Join DP. Now mAP2 + wBP2 = mAD2 + »DB2
+ (m + n) DP2 (Book II., Ex. 12). Join DC, and divide it in E,
so that (m + n) DE = j^EC. Join EP ; then (m + n) DP2 + pVC2
= («» + ») DE2 + jt?EC2 + (m + n+p) EP2; add, and mAP2 + «ΒΡ2
P
+ pVC2 = mAD2 + nDB2 +(m + n) DE* + pEC2 + (m + n + p) EP2;
butmAP2 + wBP*+pPC2 is given (hyp.); .·. mAD2 +nDB2+ Ac.,
is given; but mAD2 + »DB2 is given, and (m + n) DE2, and pEC2
is given; .*. (m + n+p)EP2, and (m + n+p) is given; .·. EP*(BOOK VI.]
EXEBCISES ON EUCLID.
243
is given, .*. EP is given, and E is a given point. Hence the locus
of P is a circle, having E for centre and EP for radius.
60. Bern.—Let P be the point. From P let fall X8 PD, PE,
PF on the sides of the Δ. Join DE, EF, FD, AP, BP, CP.
Now because the L* AEP, AFP are right, AEPF is a cyclic
-quadrilateral; then AP is the diameter of the circumscribed Θ.
Draw FG, another diameter. Join GE. Now the L FGE = FAE
<111. xxi.); but FAE is a given L, .*. FGE is a given L, and
the L FEG is given, being right; .·. the Δ FGE is given in
EF	EF
species ; hence — is given; but FG = AP ;	— is given ;
EF2
. *.	is given ; let it be equal to m, then EF2 = wAP2. In like
manner, FD2 = wBP2, and DE2 = pQY2; but EF2 + FD2 + DE2 is
given (hyp.); .·. iwAP2 + »BP2 + pQV2 is given. And hence
{Lemma) the locus of P is a circle.
61. Let the Θ W make given intercepts DD', EE'on two fixed
lines PX, PY. It is required to prove that the rectangle CG; CH
contained by the X8 from the centre C on the bisectors of the
L · formed by the lines PX, PY is given.
Bern.—From C let fall Xs CA, CB on DD', EE'. Join CD,
CE. Now AC2 + AD2 = CD2, and BC2 + BE2 = CE2; .*. AC-
+ AD2 = BC2 + BE2; .·. AD2 - BE2 = BC2 - AC2 ; but AD, BE
are the halves of DD', EE' (III. hi.), and are given (hyp.);
BC2 — AC2 is given. Now since the L8 CAP, CBP are
right, OAPB is a cyclic quadrilateral. Describe a Θ about it.
Join AB·; the line bisecting AB perpendicularly will be the dia-
meter. Let it be GH. Join GP, HP; these are the internal and
«external bisectors of the L EPD (III. xxx., Ex. 2). Join CP, CH,
B 2244
EXERCISES ON ETJCLID.
[book yi.
and let fall 1· CF, Cl on AB, OH. Now BC2 = BF2 + FC2,
and AC2 = AF2 + FC2 ; .·. BC2 - AC2 = BF2 - FA2; but BC2
—	AC2 is given; .·. BF2 — FA2 is given; that is (BF + FA) (BF
-	FA) is given; but BF + FA = AB, and BF — FA = 2 OF f
AB . OF is given; that is, AB . Cl is given. And because
the L APB is given, the ratio of AB to the diameter is given
(Dem. of Ex. 60); that is, AB : GH is given; the ratio
AB. Cl : GH . Cl is given; but AB. Cl is given; .·. GH. Cl
is given. And since the Δ8 GCH, ICH are equiangular,
GH . Cl = GC . CH. Hence GC. CH is given.
62. Let ABC be a Δ, whose base and the difference of whose
base L* is given. Draw CE, CF, the internal and external
bisectors of the vertical L. Bisect AB in D, and let fall JL· DE,
DF on CE, CF. It is required to prove that the rectangle
DE · DF is given.
Dem.—Draw BG, BH || to CF, DF. Produce CB to meet
DE produced in I. Draw IJ || to CE, and let fall a ± AK on
ID produced.
Now the L NCB= ACN, and LCJ = ICJ ; .*. the L NCJ is
right, and DFC is right; .·. DF is || to CN; .*. FCMG is a
parallelogram. Now the L ANC = NCB + CBN, and BNC
= NCA + CAN; .·. (ANC - BNC) = (NCB + CBN) - (NCA
+ CAN); but NCB = NCA ; .·. (ANC - BNC) = (CBN - CAN) ;
but (CBN — CAN) is given (hyp.), (ANC — BNC) is given,
and their sum is given; hence each is given; but DNE = BNC;
.·. DNE is given, and DEN is right; EDN is given; henceBOOK VI.]
EXERCISES ON EUCLID.
245
the line IK is given in position; .·. PB JL to JK is given in
position. And because the L QCE = ICE, and CEQ = CEI, and
OE common, EQ = El, and the L EQC = EIC; but EQC
= AQK ; .·. EIC or BIP = AQK, and AKQ = BPI, each being
right, and the side AK = BP ; .·. KQ = IP. To each add QP,
•and we have KP = QI; hence (Ax. 7) KD = QE; . KQ = DE;
. *. DE = IP; hence the figure GC = B J; but B J = BE
-(I. xiiin.);	GC = BE; hence the rectangle DC = BD; that
is, the rectangle DE . DF = BD ; but BD is a given rectangle.
Hence DE . DF is given.
63. Let ABC be the Δ. Bisect the L ACB by CD. From
A, B let fall _L8 AG, BF on CD. Produce AC, and bisect the
L BCJ by CE, meeting AB produced in E. Bisect AB in 0,
and let fall a _L OH on CD. It is required to prove that AG. FB
= OH . CE.
Dem.—Now AD : DB : : AE : EB (in., Ex. 3); hence
(Book V., Ex. 9) OD : OB : : OB : OE; that is, OD . OE
— OB2 ; but (II. ni.) OD . OE = OD2 + OD . DE, and (II. v.)246
EXERCISES ON EUCLID.
[book VI.
OB2 = AD . DB + OD2; hence OD . DE = AD . DB ; .·. AD :
OD : : DE : DB; but AD : OD :: AG : OH, and DE : DR
:: CE : FB; AG : OH :: CE : FB. And hence AG . FR
= OH . CE.
64.	The rectangle contained by the perpendiculars from the
extremities of the base on the external bisector of the vertical
angle, is equal to the rectangle contained by the internal bisector
and the perpendicular from the middle of the base on the external
bisector.
Let ACB he the Δ. Produce AC to J, and bisect the L BC J
by ECG, meeting AB produced in E. From A, B let fall _L8 AG,.
BF on EG. Bisect the L ACB by CD. Bisect AB in 0, and
let fall a _L OH on EG. It is required to prove that AG . BF
= OH . CD.
Dem.—AE. EB + OB2 = OE2 (II. vi.); hut OE2 = OD . OEL
+ DE . OE (II. ii.) ; hence AE.EB 4 0B2 = OD . 0E+ DE. OE ;
hence AE. EB = DE . OE (see Ex. 63); .*. AE : DE :: OE: EB.
Hence, by similar triangles, AG : CD :: OH : BF; .*. AG . BF
=> OH . CD.
65.	Dem.—From C let fall a _L CD on AB. Now the Δ" ACD,.
BCD, ABC are similar (viii.) ; then, if R, R', p, are the radii of
the Θ* inscribed in these Δ8, AC, BC, AB are proportional to
R, R', p ; but AC2 4 BC2 = AB2; .·. R2 4 R'2 = p2, and p*
= (s — c)2 (IV. iv., Ex. 14); that is, R2 4 R'2 = (s — c)2.
66.	Sol.—Through A, C draw two parallel lines AF, CE; and
through B, D draw two parallel lines BF, DE, meeting the \\
through A, C in F, E. Join EF, and produce it to meet AI>
in 0.
Dem.—Because BF is || to DE, the Δ8 ODE, OBF are equi-
angular ; hence OD : OB :: OE : OF ; and since the Δ8 OCEr
OAF are equiangular, OE : OF : : OC : OA, OD : OR
: : OC : OA. Hence OA . OD = OB . OC.
67.	Sol.—Let a, 5, c, d be the four sides. Find a fourth pro-
portional to (2 ab 4 2 cd), {(cP + cP) — {a2 4 52)}, and b. Let it be-
BE. Produce EB to A, so that AB = a. Erect EC _L to AE.
With B as centre, and a radius equal to b, describe a Θ, cutting-
EC in C. Join BC, AC ; and on AC describe a Δ ACD, having:
its sides CD, AD equal to c and d. ABCD is the required
quadrilateral.BOOK ΤΙ.]
EXERCISES ON EUCLID.
247
Dem.—From A let fall a _L AF on CD. Now because BE is a
fourth proportional to (2 ab + 2 cd), {(c2 + d2) - (a2 + δ2)}, and b,
we have (2ab+2cd) BE = {(<* + d2) - (a2 + b2)} b. Now AC*
* AB3 + BC2 + 2AB . BE (II. in.); that is, AC2 = a2 + b2
+ 2a. BE ; and AC2 = c2 + d2 - 2<? . DF (II. xiii.) ; .·. c2 + d*
— 2c . DF = a2 + b2 + 2a. BE; .·. d2 + d2 - (a2 + b2) = 2a . BE
-f 2 c . DF ; hence (2 ab + 2 cd) BE = (2 a . BE -f 2 c . DF) b;
.*. 2 cd. BE = 2 be. DF; .·. d. BE = £ . DF; .·. rf:DF :: 6: BE,
that is, AD : DF :: BC : BE, and the L AFD =BEC ; the A8
ADF, CBE are equiangular; .-.the L ADF = CBE. To each
add ABC, and we have the L8 ADC, ABC equal to ABC, EBC;
ADC + ABC equal two right Ls. Hence ABCD is a cyclic
quadrilateral.
68. Let A, B be the centres of the Θ8. From a point C tan-
gents CF, CE are drawn to the Θ8, so that CF : CE :: a: b.
It is required to find the locus of C.
Sol.—Join AF, BE, AC, BC, and let the radii be denoted by
R, R'. Now since CF : CE : : a : b, CF2: CE2 :: a2 : b2 ; that
is, AC2 - R2 : BC2 - R'2 :: «2 : & ; .·. b2 AC2 - b2 R2 = a2 BC2
- e2 R^; .·. b2 AC2 - a2 BC2 = Ψ R2 — a2 R'2. Join AB, and
produce it to D, and make AD : BD : : a2 : b2; then b2 AD = a2 BD.
Now, joining CD, and putting V2 for m> and a2 for ny we have248
EXERCISES ON EUCLID.
[book vi.
(Book II., Ex. 18) b2 AC2 - a2 BC2= b2 AD2 - a2 DB2 + (δ2 - a2)
CD2, and .·. (Ax. 1) b2 AD2 - a2 DB2 + {b2 - a2) CD2t= P R2
—	a2 R'2; and transposing, we get (α2 — δ2) CD2 = b2 (AD2 — R2)
-	a2 (DB2 - R'2); .·. (b2 - a2) CD2 is given ; .·. CD is given, and
the point D is given. Hence the locus of C is a circle.
69. Sol.—Describe Os about the Δ8 APD, BPC. Draw OP
a tangent to the Θ APD, meeting DA produced in 0. Now
the L OPA = PDA (III. xxxn.), and the L APB=CPD (hyp.);
.·. the L OPB = ADP -f CPD = ACP; hence OP touches the
Θ BPC. Now (III. xxxvi.) OA. OD = OP2, and OB. OC = OP2,
.·. OA. OD = OP. OC; .·. 0 is a given point (Ex. 66), and A, D
are given points; OA. OD is given; .*. OP2 is given; .·. OP
is given. Hence the locus of P is a Θ, having 0 as centre
and OP as radius.
70. If a Θ ACB be circumscribed to a Δ, and a Θ GBH be
inscribed, touching the sides AC, BC in D, F, and the circum-
scribed Θ in H. It is required to prove that CD is a fourth propor-
tional to the semi-perimeter of the Δ ABC, and the sides CA, CB.
Dezn.—Join CH, and draw HJ a tangent to the Θ ABC; at
G draw a tangent A'J to the Θ DFH. Join AH.
Because JG = JH, the L JHG = JGH; but JGH = GB CBOOK YI.]
EXERCISES ON EUCLID.
249
+ B'CG, .·. JHG = GB'C + B'CG, and AHJ = GCB' (III. xxxn.);
. GHA = GrB'C. To each, add GB'A, and we have GB'C + GB'A
= GB'A + GHA, .·. GB'A + GHA equal two right Z."; hence
GB'AH is a cyclic quadrilateral, and therefore HC. CG = AC. CB';
but HC . CG = CD2 (III. xxxvi.); .·. AC . CB' = CD2. Again,
the L CHA = A'B'C; but CHA = CBA (III. xxi.), CBA
= CB'A', and the L A'CB is common, .·. the Δ8 ABC, A'B'C are
equiangular ; and, denoting their semiperimeters by s, s', we
have (xx., Cor. 1) s : s' : : BC : B'C, .·. s : s'. CA : : BC :
B'C . CA; that is, s : s'. CA : : BC : CD2; but CD2 = s'2 (IV.
iv., Ex. 4); .*. s : s'. CA : : BC : s'2. Hence s : CA : : BC : s';
or, « : CA : : CB : CD.
71. It is an obvious modification of 70.
73. Let the sides AC, BC of the Δ ABC, circumscribed to a
given 0, be given in position, but the third side AB variable.
About ABC describe a Θ. It is required to prove that the O
about ABC touches a fixed circle.
Dem.—Describe a Θ, touching the sides AC, BC in D, H,
and the Θ about ABC in L. Let 0 be its centre. Join OD, OH.
Let fall a JL AF on BC. Draw DE || to AF, and let fall a _L OG
on DE.
Now s: CB : : CA : CD (Ex. 70); but CA : CD :: AF : DE;
therefore s : CB :: AF : DE, .·. s. DE = CB . AF = twice the area
of the Δ ABC =2 rs (IV. iv., Ex. 9), .·. DE= 2 r; but 2 r is given,
.*. DE ip given; and because the L ECD is given (hyp.), and
the L E is right, the Δ ECD is given in species, .*. the ratio ED
: DC is given; but ED is given, .·. DC is given; .*. D is a given
point.250
EXERCISES ON EUCLID.
[book VI.
Again, because the L ODC is right, and .·. = ECD + CDE,
.·. ODG = ECD. Hence ODG is given, and OGDiis right,
the Δ OGD is given in species, .·. the ratio OD: DG is given;
but OD = OH = GE, .·. the ratio EG : GD is given ; but ED is
given; .·. EG, that is OD, is given, and the point D has been
shown to he given. Hence the Θ, with 0 as centre, and OD as
radius, is a fixed Θ, and the O about ABC touches it in L.
74. Let AC, BC be the two sides given in position.
Sol.—Bisect the L ACB by FF\ In CF find a point F, such
that CF2 = CA. CB. F is one of the required points.
Dem.—Join AF, BF, and let fall a X BE on AC. Now
B
because the area of the Δ ACB is given, CA . EB is given ; and
since the L BCE is given, and the L BEC is right, the Δ BCE
is given in species, .*. the ratio CB : BE is given, .·. the ratio
CB . CA : BE . CA is given; hut CB . CA = CF2 (const.), and
BE . CA is given; .·. CF2 is given, .·. CF is given, and .·. F
is a given point. Again, because CA . CB = CF2, CA : CF : :
CF : CB, and the L ACF = BCF, .·. (vi.) the L CFA = CBF.
To each add the sum of the Z8 CFB, BCF, and we have the sum
of the L8 of the Δ CBF equal to the L8 AFB and BCF,
.·. AFB and BCF are equal to two right L8; but the L BCF
is given, .*. AFB is given. Hence the base AB subtends a con-
stant L at a given point F. In like manner it can be shown
that it subtends a constant L at F', constructed by making
CF'= FC.
75. Let ABCD be the cyclic quadrilateral. (See Diagram,
Ex. 67.)
Dem.—Draw the diagonal AC. Produce AB, and let fall the
i8 AF, CE on AB, CD.EXERCISES ON EUCLID.
251
BOOK ΎΙ.]
Now, since the sides AB, BC, CD, DA are denoted by a, b,
o, rf, we have (II. xii.) AC2 = a2 + & + 2a. BE, and (II. xrn.)
AC2 = &+& -2c. DF; .\c2 + rf2 - 2<j. DF =a2+ 52 + 2a. BE;
.·. & + rf2 - (a2 + b2) = 2λ . BE + 2e. DF; and because the
BCE, ADF are equiangular, BC : BE :: AD : DF; that
h : BE :: d : DF, .·. b . DF = d. BE, DF = - . BE ; a
2 erf
hence we have c2 + <Z2— (a2 + i2) = 2«. BE + -y · BE
{(c+<i+e-J)(e + rf-a+>)(<»+* + «-<i)(« + i-‘, + rf)t
= i	4(«J + **)2·
Hence, putting (« + } + c + d) = 2 *, and substituting, we get]
*2if«2 + «i2)-(*2+i8)}2
Again, CE2 = BC2 - BE2 = i2 —+ «*)»}
{(^2 + ^)-(«2 + y)}2)
(	4 {ab + erf)2 ?
{Mab + cdY-{(? + iP)- («2 + *2)}f
4(«i + «i)2
^--------------7ΤΤΓ~Γ777Γ'
4 (αδ + tfrf)2
ab + ed
Now AB = a, and AB . CE = 2 Δ ABC.
CE = 2ftV(s-a) (s - b)(s - c) (s -rf)
.·. ABC =
^-pr, __ ab^(s - «) (5 —
“	ab + ed
Similarly, ACD =
cd ^/(s — a) (s — b)(s- e) {$	d)_
{ab + ed)
B· >252
EXERCISES ON EUCLID.
[book ti.
Hence the quadrilateral ABCD
__ (ab + cd) y/(s — a) (s — b) (s — c) (s — d)
(ab -f cd)
= V(s — a) (s — b) (s — c) (s — d).
76. Dem.—Produce BC, C'B' to meet in A". Let fall A*
A'O', BO on A"C'.
Now AB'. BC'. CA' = A'B . B'C. C'A (Ex. 4), and AB'.BC'.CA"
PA' A'R
= A"B. B'C . C'A (Ex. 5). Divide, and we get —— = —— ;
OA A B
. ·. A"B . A'C = A"C . A'B, . ·. A"B. A'C + A"C. A'B = 2 A"B. A'C ;
that is (“ Sequel,’’ Book II., Prop, vn.), A"A'. CB = 2 A"B. A'C.
Now the Δ ABC : ABB' : : AC : AB' (i.), and ABB' : BC'K
: : AB : BC', and BC'B': A'B'C' : : BO : A'O' :: BA" : A'A";
that is, since A"A'. CB = 2 A"B. A'C :: BC : 2 A'C ; Δ ABC :
A'B'C' : : AB. BC . CA : 2 AB'. BC'. CA'.
77.	Dem.—Draw the diameter AE. Join BE, and let fall
a JL AD on BC. Now (xvn., Ex. 5) AE. AD = AB. AC;
.·. AE. AD . BC = AB. BC . CA; but AD. BC=twice the Δ ABC;
2AE.ABC = AB.BC.CA; hence (Ex. 76) ABC : A'B'C'
:: 2AE. ABC : 2AB'. BC'. CA', .·. 1: A'B'C':: AE: AB'.BC'. CA';
AE. A'B'C' = AB'. BC'. CA'; and hence
4in AB'.BC'.CA'
AE= A'B'C' '
78.	Dem.—Let the sides of the quadrilateral he denoted by
λ, 5, e, d. Now (III. xvn., Ex. 3) (a + c) = (5 + d); .·. 2 (a + e)
= (a + b + e + d). Hence, putting (a + b + e + d) — 2 s9 we haveBOOK ΤΙ.]
EXEBCISES ON EUCLID.
253»
2 (a + c) = 2 s, .*. {a + c) - s; .·. a = (s - c). Similarly, b = (s -d)r
c=(8 — a), d={s — b)\ and (Ex. 74), we have area of quadri-
lateral = V(s — a (« - b) (s — c) (s - d); .*. area = Vabcd. Hence
the square of the area = abed.
79. Dem.—Join BF, CF, BE. Let the ratio BD : AD be
denoted by m, n. Now the Δ ABC : ABE :: AC : AE(i.) :; AB:
BD (hyp.); that is, as (m + n) : in, and ABE : BDE :: (m + w) m9.
and BDE : BDF : : (m + n) : m. Multiplying together, we have-
A TIP m.3
ABC : BDF:: (m + nf : m*; hence BDF =	-V· ^ Bkfr
(m + n)
ABC
manner ECF = -r—Again (xxiii., Ex. 1), ABC : ADE
(m + n)s*
::(*» + λ)2 : mn; .*. ADE =
ABC. mn
(m + n)2 *
Now the Δ BFC = ABC - BDF - CEF - ADE = ABC
{*-
m3
(m + »)3 (m + n)2	(m+«)2
Hence the Δ BFC = twice the Δ ADE.
i-
ABC
2mn
(m + »)2*
80. Let ABCD he a quadrilateral. Join AC, BD, and bisect
them in E, F. Through E, F draw EG, FG parallel respectively
to BD, AC. Bisect AD, CD in Η, I. Join GH, GI. It is re-
quired to prove GIDH = J ABCD.
Dem.—Join HF, IF, IH. Now, because AD, BD are bisected
in H, F, HF is || to AB, and the Δ DHF = JADB (I. xl.,
Ex. 2). In like manner, DFI = £ DBC; .*. DHFI = J ABCD..
Again, HI is = to AC, and FG is || to AC; HI is || to FG ?254
EXERCISES ON EUCLID.
[book VI.
.*. (I. xxxvn.) the Δ HFI = HGI. To each add HDI, and
HDIF = HGID ; .·. HGID = | ABCD. In like maimer, if we
bisect BC in J, and join GJ, GICJ = J ABCD, &c.
81.	Dem.—Let 0, O' be the centres. Join 00', and produce it
to meet AB in G ; O'G is evidently perpendicular to AB. Com-
plete the circle on AB, and produce EC, FD to meet it again
in Ξ, I.
Now AC. DB = OG2 (xm., Ex. 5), and AD . CB = O'G2 (im,
Ex. 7); hence AC. CB. AD . DB=OG2 . O'G2; hut AC . CB =CE2,
and AD . DB = DF2 ; therefore CE2 . DF2 = OG2 . O'G2. And
hence CE . DF = OG . O'G.
82.	Let ABCDE he the inscribed regular polygon. Take any
point P in the circumference. Join PA, PB, PC, PD, PE, and
let those lines he denoted by pi, p2, pz, pi, p5· It is required to
prove that pi + pz + ps = pz + pi·
Dem.—Join BD. Let the sides of the polygon he denoted by
s, and the diagonals by d. Now, considering the polygon ABDP
formed by pi, p2, pi, we have (xvn., Ex. 13) p\d + pis = p2<f.
Similarly, we have p\d — pzs -f pis, and p$d + p^s = pid. Adding,
we get (pi + pz + pb)d = (pz + pi)d. Hence pi + pe + ps = p2 + pi·
83.	Let O be the centre of the given Θ ; P the given point;
AB any chord passing through P; PD, PE perpendiculars on theBOOK VI.]
EXERCISES ON EUCLID.
255
tangents AT, BT. It is required to prove that the sum of the
reciprocals of PD, PE is constant.
Dem.—Join OP, produce it, and from T let fall the _L TC on
OP produced. Produce AB to meet CT in H, and let fall the
±· AF, BG.
Now (“ Sequel,” Book III., Prop, xxviii.) CT is the polar
of P, and AT is the polar of A. Hence (“ Sequel,” Book III.,
Prop, xxvii.), since PD and AF are perpendiculars on the polars,
OA : OP: AF : DP;
1 OA 1
DP ” OP * AF*
In like manner,
J_ _ OB
PE” OP'
Hence, denoting the radius of the circle by r, and the distance
OP by d, we have
1	^ _ r / 1	1 \
PD+ PE~rf*\AF+BG/
Again, since P is the pole of the line GH, the line HB is cut
harmonically ; .·. HP is a harmonic mean between HA and HB;
but AF, PC, BG are proportional to HA, HP, HB ; hence PC is
a harmonic mean between AF and BG ;
2	1	11	1 r 2_
PC " AF + BG : ·'· PD + PE “ d * PC*
Hence the proposition is proved.
84. Let ABC be a cyclic quadrilateral, whose sides AB, CD,
AD pass through three collinear points E, F, G. Join BC, and256	EXERCISES ON EUCLID. [book vi.
produce it to meet EG in H. It is required to prove that H is a
fixed point.
Dem.—Through B draw BK || to EG. Join DK, and produce
it to meet EG.
Now the Z.* ADK, ABK equal two right Z8 (hyp.); but ABK
= AEG (I. xxix.), AEI and ADI are equal to two right L·;
hence AEID is a cyclic quadrilateral; .·. EG. GI = DG. GA; but
DG. GA is given ; EG. GI is given, and EG is given, .*. GI
is given; .*. I is a given point. Again, the L IDF KBC/(IIL
xxi.); but KBC = CHF; .·. IDF = CHF, and .·. the points
D, C, I, H are concyclic ; hence DF. FC = FI.FH; butDF.FC
is given; .*. FI. FH is given, and FI is given; .·. FH is given.
And hence H is a given point.
85. (1) Suppose the polygon to be a Δ. Let BCD be a Δ,
whose sides are-parallel to three given lines EF, GH, IJ; andBOOK ΤΙ.]
EXERCISES ON EUCLID.
267
let the loci of its angular points B, C, be right lines AB, AC. It
is required to prove that the locus of D is a right line.
Dem.—Join AD. Produce CA to meet EF in F.
Now the L BCA = FEA; .·. BCA is a given Z, and the
Z BAC is given, since the lines AB, AC are given in position;
hence the Δ ACB is given in species; .·. the ratio AC : CB is
given.
Similarly, the ratio BC : CD is given; the ratio AC : CD
is given, and the L ACD is given; hence the Δ ACD is given
in species ; the L CAD is given, and the line AC is given in
position; therefore the line AD is given in position. Hence the
line AD is the locus of D.
(2) Let the polygon be the quadrilateral ABCD, having its sides
parallel to four given lines, and the loci of the Z8 A, B, D right
lines.
Dem.—Let the loci of A, B meet in E. Produce DA, CB to
meet in F. Join EF.
Now AFB is a Δ, whose three sides are parallel to three given
lines, and the loci of A, B are right lines. Hence (1) the locus
of F is the line EF, which is therefore given in position.
Again, DFC is a Δ, having its sides parallel to three given
lines, and having straight lines for the loci of D and F. Hence—
(1) the locus of C is a right line. In like manner it can be proved
for a figure of any number of sides.
86. Let BAC be a A whose vertical L BAC, and its bisector
AD, are given. It is required to prove that —ί + —^ is given.
Dem.—Describe a Θ about ABC. Produce AD to meet the
circumference in E. Join EC, and let fall a X EF on AB.
s268
EXERCISES ON EUCLID.
[book ▼ .
Now AF = \ (AB 4- AC) (III. xxx., Ex. 4). And since the
L BAC is bisected by AE, FAE is a given L, and the L AFE
AF *
is right; .*. the Δ AFE is given in species; ^ is given ;
2AF - . . AB + AC .	.	_ ATk .	.	.
7=», that is, --ΓΐΓ- is given, and AD is given (hyp.);
AE	AE
.*.	is given. Again, the L ABC = AEC (III. xxi.),
AD. AE
and BAD = CAE; the Δ8 BAD, CAE are equiangular;
.·. AB: AD:: AE: AC; hence AB. AC = AD. AE; .·. AB + AC
is given; that is,
AB
AC
AB . AC + AB. AC
AB. AC
tt	1	1
is given. Hence 55 + 55
is given.
87. (1) Let the polygons be the Δ8 A'B'C', ABC. Bisect the
arcs A'B', B'C', C'A' in the points D, E, F. Join A'D, DB',
ΒΈ, EC', C'F, FA'. This hexagon is the corresponding polygon
of double the number of sides. It is required to prove that the
hexagon is a geometric mean between the Δ8 ABC, A'B'C'.
Bern.—Join AO, A'O, BO, ΒΌ, CO, C'O. Let OC intersect
A'B' in N.
Now we have the Δ OB'C : OB'D :: OC : OD (1.), and
OB'D : OB'N :: OD : ON; but OC : OD :: OD : ON; hence OB'C
: OB'D :: OB'D : OB'N; that is, the Δ OB'D is a geometric
mean between the Δ8 OB'C, OB'N; but the hexagon is six times
OB'D, ABC six times OB'C, and A'B'C' six times OB'N. Hence,
denoting the areas by P, P', *·, we see that w is a geometric mean
between P and P'.
(2) At the points D, E, F draw tangents to the Θ ; the figure,book vi.] EXERCISES ON EUCLID.
259
-whose sides are those tangents, and the parts cut off by them
from the sides AC, CB, BA, is a circumscribed polygon of double
the number of sides.
Dem.—Join OK. Now, since A'C' is parallel to OK, AO : OA'
:: AK : KC; but OA' = OE, .·. AO : EO :: AK : KC'. Again
<i.), the Δ AOC': EOC':: AO : EO, and AKE : EKC':: AK : KC';
.*. AOC' : EOC' : : AKE : EKC'. Now consider the figures
AOC', OEKC', and OEC'. AOC' is the first, OEC' the third,
and OEKC' the second; and we have shown AOC' : EOC' : :
AKE : EKC'; that is, the 1st: 3rd : : (1st - 2nd) : (2nd - 3rd);
.·. OEKC' is a harmonic mean between OEC' and AOC7; but
OEKC' is ^ of x', OEC' is ^ of x, and AOC' ^ of P'. Hence x' is
a harmonic mean between x and P'. In the same manner the
proposition may be proved for a polygon of any number of sides.
88. Lemma.—If upon the base BC of a Δ ABC two equilateral
Δ8 BCE, BCE' be described on opposite sides, and their vertices
E, E' joined to A, then (1) if S denote the area of ABC, AE'2-AE'2
= 4 S V3 ;^(2) AE2 + AE'2 = AB2 + CA2.
(1)	Dem.—Join EE',“intersecting BC in 0. Join AO, and
«draw AM, tAN perpendicular to BC, EE'. Now AE2 — AE'2
= ΕΝ2 - NE'2 = 4E0 . ON = 4 V3 . OC . ON; but OC . ON = area
of the Δ ABC = S; .♦. AE2 - AE'2 = 4 V3 . S.
(2)	AE2?+ AE'2 = 2 AO2 + 2 AE2 = 2 AO2 + 6 OC2. Again,
AB2 + AC2 = 2A02 + 2 OC2, and BC2 = 4 OC2; AB2 + BC2
+ CA2 i 2 AO2 + 6 OC2. Hence AE2 + AE'2 = AB2 + BC*
+ CA2.
Let ABC ^be the Δ ; G, H, J the circumcentres of the equi·
s 2EXERCISES ON EUCLID.
260
[book vu
lateral Δ8 constructed outwards on its sides. Join AG, AJ; BG*.
BH; CJ, CH; and GH, HJ, JG.
Now the L EBH = ABG, because each is half an L of an
equilateral Δ ; to each add HBA, and we have the L EBA
= HBG.
Again, EB2 = 3 BH2, and AB2 = 3 BG2; .·. EB : BA :: BH :
HG. Hence the Δ8 EBA, HBG are equiangular; .·. EB2:EA2
:: BH2 : HG2; but EB2 = 3BH2; . ·. EA2 = 3 GH2.
In like manner it may be proved, if G', H', J' be the circum-
centres of the equilateral Δ8 constructed inwards on the sides
of ABC, that AE'2 = 3 G'H'2. Hence AE2 - AE'2 = 3 (GH2
- G'H'2).
Again, denoting the areas of the equilateral Δ8 GHJ, G'HVU
by 2, 2', we have 2 =—2' = —3; .·. 4 V3 (2 - 2')
= 3 (GH2 - G'H'2); but 4 V3 S = AE2 - AE'2 (Lemma);
2 — 2' = S.
89.	From last demonstration we have AE2 + AE'2 = 3(GH2
+ G'H'2); but AE2 + AE'2 = AB2 +^BC2 + CA2 (Lemma);
.·. 3 (GH2 + G'H'2) = AB2 + BC2 + CA2, or the sum of the
squares of the sides of the two equilateral Δ8 GHJ, G'H'J' is
equal to the sum of the squares of the sides of the Δ ABC.
90.	(1) Let ABC be a regular polygon of three sides, the radii
of whose circumscribed and inscribed Θ8 are denoted by R, r ;
A'B'C'D'E'F' a regular polygon of the same area, and double the
number of sides, the radii of whose circumscribed and inscribed
Θ· are R', r'. It is required to prove that R' = VRr.book vi.] EXERCISES ON EUCLID.
261
Bern.—Join OA (R), ΟΆ' (R'), and let fall a X 01 (r) on AB.
Product 01 to meet the Θ in D. Join AD, B'D, 0'B\ Now (i.)
the Δ OAD : OAI : : OD : 01; that is, as R : r; hut OAI
-= O'A'B'; .·. OAD : O'A'B' : : R : r; but (xix.) OAD : O'A'B'
:: OA* : O'A'2; that is, as R* : R'*; hence R : r : : R* : R'2 ;
.·. RR'* = RV; .·. R'* = Rr. And hence R' =
(2) It is required to prove that / = -^- .
Bern.—Join OA. Let fall a X 01 on AB, and produce it to
.meet the Θ in D. Through D draw a tangent MK, and produce
OA to meet it. Through A, B draw tangents LS, MN. Bisect
the arcs AC, BC, and at the points of bisection and at C draw
tangents SR, NQ, RQ. Join OL.
Now OA : 01 : : OK : OD ; but OK : OD :	:	KL	:	LD;
..·. OA : 01 : : KL : LD ; that is, Rif:: KL : LD	;	.·.	(R-j- r):
r : : KD : LD ; and KD : LD : : Δ OKD : OLD ; .·. (R + r) :
,r:: OKD : OLD ; .·. (R + r) r: 2r2 :: OKD : 2 OLD, or OALD.
Again (xix.), r2 : R2 :: OAI : OKD. Hence, multiplying these
proportions, we get (R + r) r : 2 R2 : : OAI : OALD; but OAI :
OALD:: ABC: LMNQRS; that is, OAI: OALD :: A'B'C'D'E'F'
: LMNQRS; that is, as/* : R*; .·. (R + r) r : 2 R2 : : /* : R*;
.·. (R + r) r = 2 r'2. Hence r' =	^	·
In the same way the proposition may be proved for a polygon
-of any number of sides.
91. Bern.—Let fall a X CE on AB ; then CE = AB (hyp.).
Describe a Θ about ABC, and produce CE to meet it in 0. Let
O be the orthocentre. Cut off BF = OE. Bisect AB in D. Join262
EXERCISES ON EUCLID.
[book. YU
OD. Now since BF = OE, and AB = CE,AF = CO. Now
AF . FB + DF2 = DB2 (II. v.); that is, CO . OE + DF2=» DB2.
Again, AE . EB + DE2 = DB2; .·. CE . EG + DE2 = DB2;
.·. CE . EO + DE2 = DB2 ; (CO + OE) EO + DE2 =» DB2
.·. CO . EO + EO2 + DE2 = DB2; CO . EO + OD2 « DB2 r
.·. OD2 = DF2; .·. OD = DF, and OE = FB (const.) Hence-
OD + OE = DF + FB «= DB.
92. Let ABC be any Δ. Describe a Θ about ABC. Draw the-
diameter EF A to AB. Join CE, CF ; these are the internal
and external bisectors of the L ACB. Produce FC, AB to meet
in G. Let fall a A CH on AB ; it is evident that the Θ on DG
as diameter will he the locus of C when the base and ratio of the-
sides are given. Let 0, O' be the centres. Join OC, O'C. It is-
required to prove that AC2 — CB2 : 4 times area :: OC : O'C.
Bern.—Through 0 draw 01 |j to AB.
Now the L FOC = 2 FEC (III. xx.); but FOC = OCI :
.·. CCI= 2 FEC, and CO'D = 2 CGD. Now the L KDE = CDG.
and DEE = DCG, .·. KED = CGD, .·. OCI = CO'H, and th*
right L OIC = CHO'; .·. the Δ8 OCI, O'CH are equiangular;BOOK ΤΙ.]
EXEBCISES ON EUCLID
263
OC : O'C :: 01: CH; that ie, OC : O'C :: ΕΠΙ: CH. Again,
AC2 - CB2=AH2 - BH2 = (AH + HB) (AH - HB) = 2 AK. 2KH
=* 4 AK . KH; but area of ABC = AK . CH, .*. four times area
= 4 AK . CH ; hence AC2 — CB2 : 4 times area :: KH : CH ;
but KH : CH :: OC : O'C. Hence AC2 - CB : 4 times area ::
OC : O'C.
Lemma.—To construct a parallelogram, being given the diago-
nals and one of the angles.
Sol.—Let AB, CD be the diagonals, and E one of the angles.
On CD describe a segment CFD containing an L equal to E.
Bisect CD in 0. With G as centre, and a radius equal to J AB,
describe a Θ, cutting CFD in F. Join FG, and produce it to H.
Cut off GH = GF. Join CF, DF, CH, DH. CFDH is the
required parallelogram; for it has the L CFD = E, and its
diagonal FH = AB.
93. Let BAC be one of the L8, and AB the difference between
its diagonals.
Sol.—Erect BC 1 to AB ; to AC apply a parallelogram ACED
equal to four times the given area, and having BAC one of its
angles. Bisect BD in F. Construct a O AHIG, having one of
its diagonals, AI = AF, and the other, HG = FD, and the L BAC
for one of its angles {Lemma). AHIG is the required parallelo-
gram.
Dem.—Through I draw IJ || to HG, and let fall a X IK
on AD.
Now AP = AG2 + GI2 + 2 AG . GK (II. xn.), and (II. xm.)
IP= JG2+ GI2 - 2 JG . GK = AG2+ GI2 - 2 AG . GK ; .·. AP
— IJ2=4 AG. GK. Again, AB = AF - FD, and AD «= AF + FD,264
EXERCISES ON EUCLID.
[book vi.
.·. AB . AD = AF2 - FD2; but AF = AI, and FD = IJ,
... AF2-FD2 = AI2-IJ2; .·. AB . AD = 4 AG. GK. .Again,
since the Δ8 ABC, GKI are equiangular, we have AB : BC
: : GK : KI, AB . AD : BC . AD : : 4AG . GK : 4 AG . KI;
hence BC . AD = 4 AG . KI. Now BC . AD = □ AE, and
4 AG . KI = 4 times □ AI, .*. □ AE = 4 times □ AI; but AE
= 4 times the given area (const.). Hence AI is equal to the
given area.
94. Let A, B be the centres of two equal Θ*, C the centre of
a variable Θ, which is touched externally by A in D, and
G
internally by B in E. Let 0 be the point of intersection of two
transverse tangents PQ, KS. From C let fall Xs CK, CK' on
PQ, BS. It is required to prove that OK . OK' is constant.
Bern.—Join CB, and produce it to E. Join CA. Describe
a Θ passing through the points C, A, B. Draw the diameter
FG, passing through 0, _L to AB. Let fall a X CH on FG.BOOK VI.]
EXERCISES ON EUCLID.
265
Produce CK, and draw HM || to PQ. Let fall a _L HL on PQ.
Join BN, and let the sides BN, ON, OB of the Δ ONB be denoted
by a, b, e.
Now AC=AD+DC, and BC = CE - BE, .·. AC-BC = 2AD,
.·. AD = J (AC — BC); that is, a = ^(AC — BC); hence (nr.
Ex. 16) a1 = OF. GH; and since AB is bisected in 0, AO = OB,
and AO . OB = OF. OG, .·. OB2; that is, e2=0F. OG, and. ·. ON2;
that is, 5* = OF. OH. Now since the Δ8 ONB, HMC are equi-
angular, and that HM = LK, we have e : b :: HC : LK;
b . HC	a. OH
.·. LK = --. In like maimer OL =--:
e	e
...0K=ii=5 + e0H
Similarly,
. OK. 0K'=
c
c
b2. HC2 a2. OH2
e
a. OH
e 9
OF. OH. FH . HG—λ2. OH2
c	cl	cl
a2. OH. FH — a2. OH2 _ a2. OH (FH-OH) a2b2
c2	e2	c2 *
but a2 is constant, since a is the radius of the circle, and e2 is
a2 if2 ,
constant, because c is half the base of the Δ ACB; .·. —is
constant. Hence OK. OK' is constant.
95. Analysis.—Let ABC be a Δ whose base AB is given in
magnitude and position, and vertical L C is given in magnitude,
and P the given point in AB, whose distance CP from the vertex
is equal’ to ^ (AC + CB). Describe a Θ about the Δ ACB.
Bisect the L ACB by CD. Let fall a X DE on AB ; then, be-
cause AB and the L ACB are given, the O is given; and since266
EXERCISES ON EUCLID.
[book VI.
the L ACB is bisected by CD, the arc AB is bisected in D ;
hence D is a given point. Again, because the L ACB .is given,
its half, the L DCE, is given, and the L DEC is right; hence
the Δ DCE is given in species, .·. the ratio of DC : CE is given;
but CE = CP, because each is equal to J (AC + CB); hence the
ratio of DC : CP is given, and the points D, P are given. Hence
the locus of the point C is a circle; and therefore the point C,
where this locus cuts the Θ ACB, is given.
96. Let 0 be the middle point of the base. F, H the points of
contact of AB, AC with the Θ. Join OF, OH. Produce FO to
A
meet the Θ in G. Join CG; then, since OC = OB, and OG = OF,
and the L COG = BOF, CG is equal to BF, and the L OGC = OFB,
and is therefore a right L; hence CG is a tangent. Again, because
the L AOC is right, and OH is ± to AC, AH. HC = OH2; but AH
= AF, and HC = CG; hence AF. CG = OH2. In like manner, if
I be the point of contact of DE with the circle, DF. JG = OI2;
but OH2 = OI2, .·. AF.CG = DF. JG; hence AF : DF : : JG:
CG, .·. AD : DF :: JC : CG ; .·. AD : JC : : DF : CG or FB;
but, by similar Δ®, AD : JC :: AE: EC, .·. AE: EC :: DF : FB;
hence, componendo, AC : CE : : DB : FB; hence AC. BF
= BD. CE; but AC and BF are each given; .·. the rectangle
BD . CE is given.
97. Let AB equal half the sum of the opposite sides, and the
area equal the rectangle ABCD.
Sol.—Join BD, and in the Θ place EF = BD. At the point
F in EF make the L EFG = ABD. Join EG, and draw EH 11 to
FG. EHFG is the required trapezium.book vi.] EXERCISES ON EUCLID.	287
Dem.—From the centre 0 let fall a ± 01 on FG, and produce
it to meet EH in J. Let fall a JL EX on FG. Produce EH, and.
draw FL parallel to EK. Because EF = BD, and the L EFX
= DBA, and the right L EKF = DAB, .·. FK = AB, .·. 2 AB =*--
2 IF + 2 IK; that is = FG + EH. Again, the L8 EGF and
EHF equal two right L s, and EHF, LHF equal two right L 8;
.·. EGX = LHF, and the right L EKG = HLF, and the side
EK = FL; .·. the Δ8 EGK, FLH are equal. To each add
the figure EHFK, and EHFG = ELFK. Hence EHFG =
ABCD.
98. Analysis.—Let the polygon he the Δ ABC, whose side»
pass through the points D, E, F. Join EF, and produce it.
Through B draw BH || to EF. Join AH, and produce it to meet
EF in G. Now the L GAC = HBC (III. xxi.), and HBC = GEG
(I. mix.), .·. GEC = GAC ; .·. GAEC is a cyclic quadrilateral^268
EXERCISES ON EUCLID.
[book VI.
.·. EF.FG = AF.FC; but AF.FC is given; .·. EF.FG is
given, and EF is given; hence G is a given point Join GD,
and produce it. Through H draw HJ || to GD. Join JB. Now
the L AHJ = ABJ, and AHJ = AGI, ABJ = AGI, AGBI
is a cyclic quadrilateral; hence GD . DI = AD . DB, and is there-
fore given; but GD is given, .·. DI is given, and I is a given
point; and since JH, BH are respectively parallel to IG, EG, the
L JHB = IGE; but IGE is given, since the lines IG, EG are
given in position, .*. the L JHB is given, .·. the arc JB is given,
the chord JB is given, and we have shown that I is a given
point. Hence the question reduces to III. xv., Ex. 2. Similarly
for a polygon of any number of sides.
99. Let the Θ8 X, Y be so related that the rectangle contained
by the diameter of X, and the radius of Y, is equal to the rect-
angle contained by the segments of any chord of X passing
through the centre of Y ; then, if from any point in the circum-
ference of X we draw tangents CA, CB to Y, and join AB, it is
required to prove that AB touches Y.
Dem.—Let 0 be the centre of Y. Join CO, and produce it to
meet X in E. Through E draw EF the diameter of X. Join
BE, BF, BO. Join 0 to G, the point of contact, and let fall
a X OH on AB. Now the L EFB = ECB (III. xxi.), and the
right L EBF = OGC, the Δ* EFB, OCG are equiangular;
.·. EF : EB : : OC : OG, .·. EF. OG = EB . OC ; but (hyp.)
EF. OG = OC. OE, .·. EB = OE, .·. the L EOB = EBO, .-.EBO
= OCB + OBC; that is, EBA + ABO == OCB + OBC ; that is,
ACE + ABO = OCB + OBC ; but ACE = OCB ; .·. ABO = OBC,
and the right L OHB = OGB, and the side OB common, .·. OHBOOK ΤΙ.]
EXERCISES ON EUCLID.
269r
= OG; but OG is the radius, .*. OH is the radius; and hence
AB touches Y. Similarly, wherever we take the point in the
circumference of X, and draw tangents to Y, the base will
touch Y.
Lemma.—If any point A is taken in the circumference of a Θ,
and A joined to 0, the centre of another Θ; and if we divide AO-
in C, so that OA. OC = r2, r being the radius of 0. It is required'
to prove that the locus of C is a Θ.
Dem.—Suppose one Θ inside the other. Let O'be the centre
of the larger Θ. Produce AO to meet O' in D. Join DO',
00*, and produce 00' to meet 0' in E, F. Through C draw
CG || to DO'.
Now OA. OC = r2, and OA.OD = OE.OF, .*. OD : OC
: : OE. OF : r2; but the ratio OE. OF: r2 is given, since r is the
radius of a given Θ, and OE. OF is a given rectangle, .·. the ratio
OD : OC is given; and because the Δ® ODO', OCG are equi-
angular, OD : OC :: 00': OG, .*. the ratio 00' : OG is given ;
but 00' is given, .*. OG is given; hence G is a given point.
Again, OD : OC : : O'D : GC, .*. the ratio O'D : GC is given;
but O'D is given, since it is the radius of a given Θ; .*. GC
is given, and we have shown that G is a given point. Hence
the locus of C is a circle.
Def.—The point C is called the inverse of the point A, and the
Θ through C the inverse of the Θ through A with respect to the
Θ through B.270
EXERCISES ON EUCLID.
[book to.
100.	Let G, H, J be tbe points where Y touches the sides of
the Δ ABC. Join HO, OJ, JH. It is required to prove that
the Θ inscribed in the Δ GHJ touches a given circle.
Dem.—Join OA, OB, OC, cutting JH, HO, GJ in L, Μ, N.
Then since L, Μ, N are the middle points of the sides of the
Δ GHJ, the Θ through these points will be the nine-points Θ
of GHJ, and will (Ex. 31) touch its inscribed Θ. Again, the Θ.
through LMN will evidently be the inverse of X with respect to Y
(Lemma), and will be a given Θ. Hence the inscribed Θ of the
Δ GHJ touches a given circle.
101.	See “Sequel,” Book VI., Prop, xii., Sect, iv., Cor. 2.
102.	Sol.—Let A, B, C be the given points; join them, and
on AB, AC describe segments of Θ8 containing L8 equal to one-
third of four right LB. Let them intersect in D. D is the point
required.
Dem.—Join AD, BD, CD ; and through D draw EF J_ to CD.
Now the L ADC = BDC, and EDC = FDC; .·. ADE = BDF;
hence (“Sequel,” Book I., Prop, xxi., Cor. 1) the sum of AD
and DB is a minimum; and CD, being a _L, is less than any
other line from C to EF. Hence the sum of the lines AD, BD,
CD is a minimum.
103.	Let AB, AC be the tangents, and 0 the centre. Join BC.
Join AO, cutting BC in F. Through A draw AD, cutting the Θ
in E, D, and BC in H. It is required to prove that AD is divided
harmonically.
Dem.—Join OC, OD, OE, OF. Join DF, and produce it.
Now (I. XLvn., Ex. 1) AO. OF = OC2 = OD2; AO : OD
:: OD : OF, and the L AOD common; hence (vl) the L ADOBOOK VI.]
EXERCISES ON EUCLID.
271
= OFD; but because OD = OE, the L ODE = OED; OFD
= OED;	OFED is a cyclic quadrilateral; .·. the L8 EDO and
EFO equal two right L8; but the L8 EFO, EFA equal two right
Z·; EFA = EDO, and EDO = OFD; .·. EFA = OFD, and
AFB = OFB; .·. DFH = AFH; hence the L EFD is bisected
internally, and the L OFD = AFG, and OFD = EFA; .·. EFA
= AFG. Hence EFD is bisected externally, and therefore ED
(in., Ex. 3) is divided harmonically in the points H, A.
104. Let A, B, C, D be the four points, and EFGH the given
quadrilateral. It is required to construct a quadrilateral similar
to EFGH whose sides shall pass through the points A, B, C, D.
Sol.—Join AB, and on it describe a segment AIB, containing
an angle equal to FEH. Join CD, and on it describe a segment
CJD, containing an angle equal to FGU. Join EG. At the
point A in ABfmake* the L BAL « FEG; and at the point
A272
EXERCISES ON EUCLID.
[book IV.
D in DC make the L CDK = EGF. Join KL, and produce it to
meet the Θ® in I, J. Join IA, IB, and produce. Join JC, JD,
and produce. INJM is the required quadrilateral.
Dem.—For the L BIL = BAL = FEG, and the L CJK = CDK
=3 EGF; .*. the Δ® INJ, EFG are similar. And because the
L BIA = FEH, .·. MIJ = HEG. Similarly, MJI = HGE,
the Δ8 MIJ, HEJ are similar. Hence the quadrilaterals are
similar.
105. Let ABCD be the given quadrilateral, and EF, FG, GH,
HE the given lines.
Sol.—Construct the quadrilateral E'F'G'H' similar to EFGH,
whose sides pass through the points A, B, C, D (103). Divide
EF in A', so that EA' : A'F : : E'A : AF'; and divide EH in B%
so that EB' : B'H :: E'B : BH'; and similarly for the other sides.
Join A'B', B'C', C'D', D'A'. It is evident that A'B'C'D' is simi-
lar to ABCD.
106. Let AB be the base, and DCE the difference of the base
angles.
Sol.—Bisect AB in F. Draw BG, making the L FBG = DCE,
and the rectangle FB . BG equal to the rectangle under the sides.
Join FG. Bisect the L BFG by FH, and make FH a mean pro-
portional between FG and FB. Join AH, BH. ABH is the re-
quired triangle.
Dem.—Produce HF to I, so that IF = FH. Through G draw
GJ || to HI, and produce BA to meet it. Join IJ, IB, GH. Now
(I. xxix.) the L HFB = GJF, and GFH = FGJ; but HFB = GFH
(const.), .*. GJF = FGJ, and .·. FG=FJ. NowtheAGHI=JIH.
To each add HGJ, and we have GHI + HGJ = JIH + HGJ.;
JIH + HGJ are equal to two right L8; hence HIJG is a■BOOH ΥΙ.]
EXERCISES ON EUCLID.
273
^cyclic quadrilateral. And since FG. FB = FH3 (const.), and
FG = FJ, and FH3 = FH. FI, .·. FJ. FB = FM. FI; .·. JIBH
is a cyclic quadrilateral. Hence the five points F, I, B, H, G are
in a circle. Now the L HBG = IBJ; but IBJ = BAH; .·. HBG
= BAH; .*. FBG, that is DCE, is the difference between HAB
and HBA. Again, the Δ8 IBF, GBH are equiangular; .-.IB:
BF :: GB : BH; .·. IB. BH = BF.BG; that is, AH.BH
= BF . BG. This construction is due to Hamilton.
107. Let the line EF produced meet AB produced in G; cut
off GH = EG. Join EH, and let fall the JL EN. Now since
(hyp.) the L AEF = EAB, the Δ AEG is isosceles; .*. AG = EG,
and EG = GH; hence the L AEH is right, .*. AN . NH = EN2 :
but EN = 2 AN ; since CD is bisected, .·. NH = 2 EN = 2 AB,
AH =
5AB
2 :
hence AG =
5AB
·. BG*
AB
4 *
Hence EC
= 2 BG ; .*. CF = 2 FB.
108. Let C be a fixed point in the diameter AB; DE a chord
passing through C. Join AD, AE. At B draw FG a tangent to
the Θ, and produce AD, AE to meet it in F and G. It is
-required to prove that BF . BG is constant.
τ274
EXERCISES OH EUCLID.
[book vu
Dem.—Through C draw HI || to BG, meeting AF, AG in
I, H. Join BE, BH. How the L BCH is right, and £EH ie
right, CBEH is a cyclic quadrilateral, the L BEC = BHC;
but BEC = BAD (III. xxi.), .·. BHC = BAD, .·. the four
points B, H, A, I are coneyclic; hence IC . CH = AC. CB;
and because the Δ8 ACI, ABF are equiangular, AC : AB :: IC
: BF; and since the Δ8 ACH, ABG are equiangular, we have
AC : AB : : CH : BG, .·. AC2 : AB2 :: IC. CH : BF.BG;
that is, AC2 : AB2 :: AC.CB : BF.BG; but the first threw
terms of this proportion are constant. Hence the fourth,
BF. BG, is constant.
109. Let 0, O' be the centres of the Θ®, and C the point of
contact.
Dem.—Join 00r, and produce it * 00' must pass through C~ηοοχ νι.]
EXERCISES ON EUCLID.
275
Let E, D be the other points in which it meets the Θ*. Join
AD, BD, AE, BE, and let AE, BE meet B'D, A'D in F, G.
Now each of the L* CBE, CB'D is right (III. xxxi.); .·. BG
is || to B'D, and BB' =■ GD. In like manner AA' = GE, .·. AA'2
+ BB'2 = GE2 + GD2 = DE2; but DE2 is constant, since it is
•equal to the square of the difference of the diameters. Hence
AA'2 + BB'2 is constant.
110.	Let d be the point of intersection.
Bern.—Join aa\ bb*; those lines must pass respectively through
the centres o', o (hyp.). Now the sides of the Δ bdb' are cut by
cd in the points c, o, d; hence (vi., Ex. 6),
dc bo b’c
eb * ob'' ed
but bo— ob*;
de b’c'
eb * cd
dd
eb " Vf
In like manner, from the Δ ada\ we get
dc	dd ca	a’cf
ca	ard* * * eb	b’c *
that is, ac: eb : : a'c' : b'cr. Hence ab : eb : : a'b* : b'd.
111.	“ Sequel,” Diagram, p. 32. By “ Sequel,” Prop, m,
<!or. 3, p. 32, we have AB. QR = EP2. Similarly AB, mul-
tiplied by the diameter of the G touching EP, the semicircle
ACB, and the semicircle on AP as diameter, is equal to EP2.
Hence the Θβ are equal.
112.	Let P be the given point, AB the chord, and CA, CB the
tangents.
Bern.—Let 0 be the centre. Join OA, OB, OP, OC, PE.
Bisect OP in D. Join DE.
Now because OA = OB, OC common, and the base CA = CB,
the L AOC = BOC; and since AO » BO, OE common, and276
EXERCISES ON EUCLID.
[boos. vi-
the L AOE = BOE, the base AE = BE. Now AO2 = AE*
+ E02; but (I. xii., Ex. 2) the lines AE, EB, EP are equal*
AO2 = OE2 + EP2 = 2 OD2 + 2 DE2 (II. x., Ex. 2);
but AO2 is given, .*. 2 OD2 + 2DE2 is given, and 2 OD2 ie
given, since OP is given, .·. DE is given, and D is a fixed point-
Hence the locus of E is a O, having D as centre, and DE as
radius. Now (I. xnvn., Ex. 1) CO . OE = OA2 — R2; .*. C, E
are inverse points with respect to the Θ ABF, and it has been
shown that the locus of E is a Θ. Hence the locus of C is
a circle.
113. Let ABC be a Δ, 0 the centre of the inscribed O, and
D the point in which the Θ touches AB. Join CO, and produce-HOOK VI.]
EXERCISES ON EUCLID.
27 r
it. CO is the bisector of the L ACB. From A let fall a X AE
on CF, and produce it to meet CB in H, and from B let fall
a X BF on CF. It is required to prove that AD . DB = AE . BF.
Bern.—Join OD. Bisect AB in G. Join EG, and produce it
to meet BF in I. Join FG. Let the sides of the Δ ABC he
denoted by a, bt c, and we have (IV. nr., Ex. 2) BD = (s - b),
AD = (s — a), .·. BD — AD; that is, 2GD = (a - δ). Now
since the L ECH = ECA, and the right L CEH = CEA, and the
side EC common, .·. CH = CA, .·. CB — CA = BH; that is
(a — b) or 2 GD = BH = 2 GE; .*. GD = GE. In like manner
GD = GF, and GD = GI; hence the lines GE, GD, GF, GI are
equal, and the Θ, with G as centre, and GD as radius, will pass
through E, F, I. Let it cut BD in X. Now (III. xxxvi.)
BD · BE = BF. BI. But since AG = GB, and DG = GE, AD
= EB. Also BI = HE = AE. Hence BD . AD = BF. AB.
114.	See “ Sequel,” Book VI., Prop, x., Sect, i., Cor. 1.
115.	See “Sequel,” Book VI., Prop, x., Sect, i., Cor. 2.
116.	Analysis.—Let P he the required point. Join AP, BP,
CP, DP. Now (hyp.) the L APC is bisected, .*. AB : BC
:: AP : CP (hi.) ; but the ratio AB : BC is given, .·. AP : CP
P
is given, and the base AC is given; hence (hi., Ex. 6) the locus
of P is a circle. Similarly for the Δ BPD, the locus of P is
another Θ. Hence the point in which these Θ8 intersect is
the point required.
117.	Let ABC be a Δ whose sides are denoted by a, b, c.
Bisect the L ACB by CD, and let CD be denoted by λ. Now (in.)
we have a : b :: BD : DA; .·. (a + b) : b :: BA : AD ; that is
he	&c
(a + b) j b :: c : AD; .·. AD =------r. Similarly, BD = --------7,
a + b	a + 0
.·. BD . DA = ,	; but ab = BD. DA + CD3 (xyii., Ex, 1);
(a + ol)278
EXERCISES ON EUCLID.
[book VI-
ab(A	abc2
.·. eb = ;--77^ + CD2; that is, ab — ;--— = CD2; that is,
(a + b)2	’	(a + b)2	\
«4 j1 - (ΓΤψ| = 01)2; hence ^ j(-(+a^-|
_ ab (a 4- b + c) (a + b - c) _±ab. s. s — c
(a + b)2	~	(β'+'ft)*	*
In like manner, denoting the bisectors of the angles A, B by a. £
respectively, we have
4 be. s .s — a
= {b + cf
, and β2=
4	. * . * — b
hence
2^2 2_64&2fl2e2 . a2 (s . 8 — g.8 — b. s — c) _ 64 g262c2.«2.(area)2
Hence,
+ ^)2 (b -f c)2 (c + <
8 abe. s. area
(c + af ’
64 a2b2c2.
[a+b)2(b+c)2{c+ a)2
afiy
(a + b) {b + c) (c + a)'
118. Let A a\ B b\ Cc' be the bisectors of the L%; then (in.) we
have c : a :: AV : VC; .·. e : e + a :: Ab* : b; .·. Abr  --.
e +1*
In like manner, Be =	, and Ca' = ^— ; .·. Ab'.BS.Ca?
a + o	b + e
a2b2c2
~ (a + b) {b + c) (c + a)'
119. Let ABC he a Δ. Draw any three lines Aa, Bb, Ce9
intersecting in D. Describe a Θ, passing through the pointsbook τι.] EXERCISES ON EUCLID.
279
«, b, e, and cutting the sides of the Δ ABC in A', B', C'. It is
required to prove that the lines A A', BB', CC' are concurrent.
Hem.—Now we have Ab. AB' = Ac. AC', Be. BC' = Ba. BA*,
and Ca . CA' = Cb. CB'; (Ab. Be. Ca) (AB'. BC'. CA') =
(aB. bC . cA) (A'B. B'C . C'A) ; hut Ab. Be. Ca = aB . bC. cA
(Ex. 4); .*. AB'. BC'. CA' = A'B. B'C. C'A. And hence the lines
AA', BB', CC' are concurrent.
120. Dem.—Describe a Θ about ABC. Let 0 be the centre.
Join CO, and produce it to meet the circumference in D. Join
DA, DB, and from A' let fall a X A'E on AB.
Now if we denote the sides by a% b, c, and the parts A'B, B'C,
C'A, by x, y, z, we have (a — x) (b — y) (c — z) = AB'. BC'. CA',
and xyz = A'B. B'C . C'A; .·. abc — (abz + bex *f cay) + ayz -f- bzx
+ exy = AB'. BC'.CA' + A'B. B'C . C'A. Again, since the Δδ
BA'E, ACD are equiangular, we have BA': A'E :: CD : CA ;
that is (denoting CD by 5), x : A'E :: δ : b; .*. bx = δ. A'E;
bx . BC' = δ . A'E . BC' = δ. 2 A'BC'; that is, bx (e - z) =
δ. 2 A'BC'; . *. {bex — bxz) = δ. 2 A'BC'. In like manner {cay — cyx)
^ δ. 2B'CA', and (abz - axy) = δ.2CAB', and “Sequel,*’
Book VI., Prop, v., Sect, i.) abe = δ. 2 ABC; abe — (hex + cay
+ abz) + (ayz + bzx + exy) = δ. 2 A'B'C'. Hence AB'. BC'. CA'
+ A'B. B'C . C'A = δ 2 A'B C'.
121. Let A, B, C be the fixed points, and the given ratio that
of 2:1.
Sol.—Take any point 0. Join OA, and produce it to D, so that
OA = 2 AD. Join DB, and produce to E until DB = 2 BE. Join580
EXERCISES ON EUCLID.
[book vi.
EC, and produce it to F, so that EC = 2 CF. Join OF, and divide
it in G, so that OG = 8 FG. Join GA, and produce it; and
through D draw DH || to OG. Join HB, and produce it; and
through E draw EJ jj to DH. Join JC, GC. GHJ is the
required Δ.
Dem.—The Δ8 OAG, DAH are equiangular; .·. OA : AD
:: OG: DH ; .·. OG=2DH; but OG = 8GF; DH = 4GF.
Similarly, from the Δ8 BDH, BEJ we get DH = 2 JE ; .*. JE
*= 2 GF, and EC = 2 CF (const.), and the L JEC = GFC ;
hence (vi.) th L JCE = GCF, and therefore JC and GC are
in the same straight line ; and evidently the sides are divided
in the points A, B, C in the given ratio. Similarly for any
polygon of an odd number of sides, and for any given ratio.
122. Let ABCD be a cyclic quadrilateral whose third diagonal
EF is a chord of another given O. Bisect EF in G. It is re-
quired to prove that the locus of G is a circle.
Dem.—Let 0, O' be the centres. Join OG, O'G, ΟΈ.
From E, F draw tangents EH, FK to 0. Join OH, OK,
EO, FO, 00'.
Now 4 EG2 -f 4 GO'2 = 4 EO'2 ; that is, EF2 + 4 GO'2 = 4 EO'2;
but EF2 = EH2 + FK2 (III., Ex. 19), and OH2 + OK2 = 2 OH2.
Adding, we get EO2 + OF2 + 4 GO'2 = 4 EO2 + 2 OH2; that is
(II. x., Ex. 2), 2 EG2 + 2 GO2 + 4 GO'2 = 4 EO'2 + 2 OH2, and
2 EG2 + 2 GO'2 = 2 EO'2. Subtracting, we have 2 GO2 + 2 GO'2
= 2EO'2 + 2OH2; .·. GO2 + GO'2 = EO'2 + OH2; but EO'2
and OH2 are given ; .·. GO2 + GO'2 is given. Therefore
OGO' is a Δ whose base is given, and the sum of theBOOK VI.]
EXERCISES ON EUCLID.
261
squares of its sides. Hence (II. x., Ex. 3) the locus of G is
a circle.
123. Let X, Y be the Θ», and let AB, a chord of X, touch Y
in C. Bisect the arc AB in D. Join DC. It is required to prove
that DC passes through a given point.
Dem.—Join 00', and produce 00', DC to meet in E. E is
the given point. Join OA, OB. Now OA = OB, OF common,
and the L AOF = BOF ; hence the L AFO = BFO; the
L AFO is right, and FCO' is right; .*. 0D is j| to O'C; hence
u282
EXERCISES ON EUCLID.
[book VI.
(n.) the Δ8 DOE, CO'E are equiangular; .·. DO : CO':: OE
: O'E; hence the ratio OE : ΟΈ is given, .·. the ratio 00': O'E
is given; but 00' is given, .*. O'E is given, and 0' is a given
point; .·. E' is a given point.
124. Let ABC be a given triangle. From a point P, within it,
let fall Xs PD, PE, PF on the sides BC, CA, AB. Join DE, EF,
FD, and let the area of DEF be given. It is required to prove
that the locus of P is a circle.
Dem.—Join AP, BP, CP. Because each of the L8 AEP,
AFP is right, AFPE is a cyclic quadrilateral. Bisect AP
in G. G is the centre of the Θ. Similarly, BDPF, CDPE
are cyclic quadrilaterals; and H, J, the middle points of BP,
CP, are the centres of their circumscribed Θ8. Join DH, HF,
FG, GE, EJ, JD. Produce FG, and let fall a X EK on it.
Because AG = GP, the Δ AGF = PGF ; .*. AFP = 2PGF.
In like manner, AEP = 2 EGP; hence the quadrilateral AFEP
= 2EGFP. Similarly, BFPD = 2 FHDP, and CDPE = 2PDJE;
hence the area of the figure EGFHD J is given; but the area of
FDE is given (hyp.); hence the sums of the areas EGF, FHD,
DJE is given. Again, the L FGE = 2 FAE (III. xx), .·, the
L FGE is given; .·. the L KGE is given, aud the L GKE is right;
hence the Δ EGK is given in species; .·. the ratio EG : EK is
given; .·. the ratio EG . FG : EK . FG is given ; but EK. FG
= 2 Δ EGF, and EG. FG = FG2; .*. the Δ EGF has a given
ratio to FG2, and FG2 has a given ratio to AP2, since AP = 2 FG;
EGF
.·.	- is given. Suppose it equal to l; hence EGF = l. AP2.
In like manner, FHD = m. BP2, and DJE = n. CP2 ; but we have
shown that the sum of EGF, FHD, DJE is given; hence l. AP2
+ m. BP2 + ». CP2 is given. And hence {Lemma to Ex. 60) the
locus of P is a circle. Similarly, the proposition may be proved
for a figure of any number of sides.Price 4/6, post free·']
THE FIRST BIX BOOKS
OF THE
ELEMENTS OF EUCLID,
With Copious Annotations and Numerous Exercises.
BY
JOHN CASEY, LL.D., P.E.S.,
Feflow of the Royal University of Ireland; Vice-President, Royal
Irish Academy; &c. &c.
Dublin: Hodges, Figgis, & Co. Ijondon: Longmans, Qreen, Λ Co.
OPINIONS OF THE WORK.
The following are a few of the Opinions received by
Dr. Casey on this Work:—
“ Teachers no longer need be at a loss when asked which of
the numerous4 Euclids’ they recommend to learners. Dr. Casey’s
will, we presume, supersede all others.”—The Dublin Evening
Mail.
44 Dr. Casey’s work is one of the best and most complete
treatises on Elementary Geometry we haye seen. The annota-
tions on the several propositions are specially valuable to stu-
dents.”—The Northern Whig.
44 His long and successful experience as a teacher has eminently
qualified Dr. Casey for the task which he has undertaken. . . .
We can unhesitatingly say that this is the best edition of Euclid
that has been yet offered to the public.”—The Freeman’s
Journal.
From the Key. B. Townsend, F.T.C.D., &c.
“ I have no doubt whatever of the general adoption of your
work through all the schools of Ireland immediately, and of
England also before very long.”( 2 )
From George Francis FitzGerald, Esq., F.T.C.D.
“ Your work on Euclid seems admirable, and is a great im-
provement in most ways on its predecessors. It is a great thing
to call the attention of students to the innumerable variations in
statement and simple deductions from propositions. ... I should
have preferred some modification of Euclid to a reproduction, but
I suppose people cannot be got to agree to any.”
From H. J. Cooke, Esq., The Academy, Banbridge.
“ In the clearness, neatness, and variety of demonstrations, it
is far superior to any text-book yet published, whilst the exercises
are all that could be desired.”
From James A. Poole, M.A., 29, Harcourt-street, Dublin.
“ This work proves that Irish Scholars can produce Class-books
which even the Head Masters of English Schools will feel it a
duty to introduce into their establishments.”
From Professor Leebody, Magee College, Londonderry.
“ So far as I have had time to examine it, it seems to me a
very valuable addition to our text-books of Elementary Geometry,
and a most suitable introduction to the * Sequel to Euclid,’ which
I have found an admirable book for class teaching.”
From Mrs. Bryant, F.C.P., Principal of the North London
Collegiate School for Girls.
“Iam heartily glad to welcome this work as a substitute for
the much less elegant text-hooks in vogue here. I have begun
to use it already with some of my classes, and find that the
arrangement of exercises after each proposition works admirably.”
From the Bev. J. E. Beffe, French College, Blackrock.
“ I am sure you will soon be obliged to prepare a Second
Edition. I have ordered fifty copies more of the Euclid (this
makes 250 copies for the French College). They all like the book
here.”
From the Nottingham Guardian.
“ The edition of the First Six Books of Euclid by Dr. John
Casey is a particularly useful and able work. . . . The illus-
trative exercises and problems are exceedingly numerous, and
have been selected with great care. Dr. Casey has done an
undoubted service to teachers in preparing an edition of Euclid
adapted to the development of the Geometry of the present day.”( 8 )
From the Leeds Mercury.
“ There is a simplicity and neatness of style in the solution of
the problems which will be of great assistance to the students in
mastering them. ... At the end of each proposition there is an
examination paper upon it, with deductions and other proposi-
tions, by means of which the student is at once enabled to test
himself whether he has fully grasped the principles involved....
Dr. Casey brings at once the student face to face with the diffi-
culties to be encountered, and trains him, stage by stage, to solve
them.,,
From the Practical Teacher.
“The preface states that this book ‘is intended to supply a
want much felt by Teachers at the present day—the production
of a work which, while giving the unrivalled original in all its
integrity, would also contain the modem conceptions and de-
velopments of the portion of Geometry over which the elements
extend.’
“ The book is all, and more than all, it professes to be. . . . The
propositions suggested are such as will be found to have most
important applications, and the methods of proof are both simple
and elegant. We know no book which, within so moderate
a compass, puts the student in possession of such valuable results.
‘‘ The exercises left for solution are such as will repay patient
study, and those whose solution are given in the book itself will
suggest the methods by which the others are to be demonstrated.
We recommend everyone who wants good exercises in Geometry
to get the book, and study it for themselves.”
From the Educational Times.
“The editor has been very happy in some of the changes he
has made. The combination of the general and particular enun
ciations of each proposition into one is good; and the shortening
of the proofs, by omitting the repetitions so common in Euclid, is
another improvement. The use of the contra-positive of a proved
theorem is introduced with advantage, in place of the redttctio ad
absurdum ; while the alternative (or, in some cases, substituted)
proofs are numerous, many of them being not only elegant but
eminently suggestive. The notes at the end of the book are of
great interest, and much of the matter is not easily accessible.
The collection of exercises, * of which there are nearly eight
hundred^’ is another feature which will commend the book to
teachers. To sum up, we think that this work ought to be read
by every teacher of Geometry ; and we make bold to say that no
one can study it without gaining valuable information, and still
more valuable suggestions.”( 4 )
From the Journal of Education, Sept. 1, 1883.
“ In the text of the propositions, the author has adhered, in all
hut a few instances, to the substance of Euclid’s demonstrations,
without, however, giving way to a slavish following of his occa-
sional verbiage and redundance. The use of letters in brackets
in the enunciations eludes the necessity of giving a second or
particular enunciation, and can do no harm. Hints of other
proofs are often given in small type at the end of a proposition,
and, where necessary, short explanations. The definitions are
also carefully annotated. The theory of proportion, Book Y., is
given in an algebraical form. This hook has always appeared to
us an exquisitely subtle example of Greek mathematical logic,
but the subject can be made infinitely simpler and shorter by a
little algebra, and naturally the more difficult method has yielded
place to the less. It is not studied in schools, it is not asked for
even in the Cambridge Tripos ; a few years ago, it still survived
in one of the College Examinations at St. John’s, but whether
the reforming spirit which is dominant there has left it, we do
not know. The book contains a very large body of riders and
independent geometrical problems. The simpler of these are
given in immediate connexion with the propositions to which
they naturally attach; the more difficult are given in collections
at the end of each book. Some of these are solved in the hook,
and these include many well-known theorems, properties of ortho-
centre, of nine-point circle, &c. In every way this edition of
Euclid is deserving of commendation. We would also express a
hope that everyone who uses this book will afterwards read the
same author’s ‘ Sequel to Euclid,’ where he will find an excellent
account of more modem Geometry.”FOURTH EDITION, Revised and Enlarged—3/6, cloth.
A SEQUEL
TO THE	^
FIBST SIX BOOKS OF THE ELEMENTS
OF EUCLID.
BY
JOHN CASEY, LL.D., F.B.S.,
Fellow of the Royal University of Ireland; Vice-President, Royal
Irish Academy; &c. &c.
Dublin: Hodges, Figgis, & Co. London: Longmans, Green, & Co
EXTRACTS FROM CRITICAL NOTICES.
“Nature,” April 17, 1884.
“ We have noticed (‘ Nature,* vol. xxiv., p. 52; vol. xxvi.,
p. 219) two previous editions of this book, and are glad to find
that our favourable opinion of it has been so convincingly in-
dorsed by teachers and students in general. The novelty of this
edition is a Supplement of Additional Propositions and Exercises.
This contains an elegant mode of obtaining the circle tangential
to three given circles by the methods of false positions, construc-
tions for a quadrilateral, and a full account—for the first time in
a text-book—of the Brocard, triplicate ratio, and (what the author
proposes to call) the cosine circles. Dr. Casey has collected toge-
ther very many properties of these circles, and, as usual with
him, has added several beautiful results of his own. He has
done excellent service in introducing the circles to the notice of
English students. ... We only need say we hope that this
edition may meet with as much acceptance as its predecessors,
it deserves greater acceptance.**( 6 )
The “Mathematical Magazine,” Erie, Pennsylvania.
“Dr. Casey, an eminent Professor of the Higher Mathematics
and Mathematical Physics in the Catholic University of Ireland,
has just brought out a second edition of his unique ‘ Sequel to
the First Six Books of Euclid,’ in which he has contrived to
arrange and to pack more geometrical gems than have appeared
in any single text-hook since the days of the self-taught Thomas
Simpson. ‘ The principles of Modem Geometry contained in the
work are, in the present state of Science, indispensable in Pure
and Applied Mathematics, and in Mathematical Physics; and it
is important that the student should become early acquainted
with them.’
“ Eleven of the sixteen sections into which the work is divided
exhibit most excellent specimens of geometrical reasoning and
research. These will be found to furnish very neat models for
systematic methods of study. The other five sections contain
261 choice problems for solution. Here the earnest student will
find all that he needs to bring himself abreast with the amazing
developments that are being made almost daily in the vast regions
of Pure and Applied Geometry. On pp. 152 and 153 there is an
elegant solution of the celebrated Malfatti’s Problem.
“ As our space is limited, we earnestly advise every lover of
the ‘ Bright Seraphic Truth ’ and every friend of the ‘ Mathe-
matical Magazine ’ to procure this invaluable book without
delay.”
The “Schoolmaster.”
“ This book contains a large number of elementary geometrical
propositions not given in Euclid, which are required by every
student of Mathematics. Here are such propositiosns as that^Ae
three bisectors of the sides of a triangle are concurrent, needed in
determining the position of the centre of gravity of a triangle;
propositions in the circle needed in Practical Geometry and Me-
chanics ; properties of the centres of similitudes, and the theoriesof inversion and reciprocations so useful in cerain electrical ques-
tions. The proofs are always neat, and in many cases exceedingly
elegant.”
The “Educational Times.”
4* We have certainly seen nowhere so good an introduction to
Modem Geometry, or so copious a collection of those elementary
propositions not given by Euclid, but which are absolutely indis-
pensable for every student who intends to proceed to the study of
the Higher Mathematics. The style and general get up of the
book are, in every way, worthy of the ‘ Dublin University Press
Series,’ to which it belongs.”
The School Guardian.”
“This book is a well-devised and useful work. It consists of
propositions supplementary to those of the first six books of
Euclid, and a series of carefully arranged exercises which follow
each section. More than half the book is devoted to the Sixth
Book of Euclid, the chapters on the 1 Theory of Inversion ’ and
on the ‘ Poles and Polars ’ being especially good. Its method
skilfully combines the methods of old and modem Geometry; and
a student well acquainted with its subject-matter would be fairly
equipped with the geometrical knowledge he would require for
the study of any branch of physical science.”
The “ Practical Teacher.”
“ Professor Casey’s aim has been to collect within reasonable
compass all those propositions of Modem Geometry to which
reference is often made, but which are as yet embodied nowhere*
... We can unreservedly give the highest praise to the matter
of the book. In most cases the proofs are extraordinarily neat.
. . . The notes to the Sixth Book are the most satisfactory.
Feuerbach’s Theorem (the nine-points circle touches inscribed
and escribed circles) is favoured with two or three proofs, all of
which are elegant. Dr. Hart’s extension of it is extremely well( 8 )
proved. . . .We shall have given sufficient commendation
to the book when we say, that the proofs of these (Idalfatti’s
Problem, and Miquel’s Theorem), and equally complex problems,
which we used to shudder to attack, even by the powerful wea-
pons of analysis, are easily and triumphantly accomplished by
Pure Geometry.
“ After showing what great results this book has accomplished
in the minimum of space, it is almost superfluous to say more.
Our author is almost alone in the field, and for the present need
scarcely fear rivals.”
The c‘Academy.”
“ Dr. Casey is an accomplished geometer, and this little book
is worthy of his reputation. It is well adapted for use in the
higher forms of our schools. It is a good introduction to the
larger works of Chasles, Salmon, and Townsend. It contains
both a text and numerous examples.”
“ Journal of Education.”
“ Dr. Casey’s4 Sequel to Euclid’ will be found a most valuable
work to any student who has thoroughly mastered Euclid, and
imbibed a real taste for geometrical reasoning. .	.	. The
higher methods of pure geometrical demonstration, which form
by far the larger and more important portion, are admirable; the
propositions are for the most part extremely well given, and will
amply repay a careful perusal to advanced students.”Recently published, Crown 8vo, 350 pp.. cloth, 7/6-
A TREATISE
ON THE
ANALYTICAL GEOMETRY OF THE POINT, LINE,
CIRCLE, AND CONIC SECTIONS.
CONTAINING
Jkcmmt'jof its must %αεχά (Sktensians.
WITS NUMEROUS EXAMPLES.
τ,τ
JOHN CASEY, LL.D., F.R.S.,
Fellow of the Royal University of Ireland; Vice-President, Royal
Irish Academy; &c. &c.
Dublin: Hodges, Figgis, & Co. London: Longmans, Green, & Co.
OPINIONS OF THE WORK.
From George Gabriel Stokes, President of the Royal Society.
“ I write to thank you for yonr kindness in sending me your
hook on the ‘Analytical Geometry of the Point, Line, Circle,
and the Conic Sections.’
“I have as yet only dipped into it, being for the moment very
much occupied. Of course from the nature of the hook there is
much that is elementary in it; still I see there is much which I
should do well to study.”
From Professor Cayley, Cambridge.
‘‘I have to thank you very much for the copy you kindly
&nt me of your treatise on ‘Analytical Geometry.’ I am glad to
see united together so many of your elegant investigations in this
subject.”
From Μ. H. Brocard, Capetaine du Genie a Montpellier.
“ J’ai en hier l’agreable surprise sur laquelle je me complais
d’ailleurs depuis un certain temps de recevoir votre nouveau
‘ Traite de Geometric Analytique.* . . .
“ Je me suis emerveille du soin que vous avez apporte a la redac-
tion de cet ouvrage, et je forme des aujourd’hui les vceux les plus
smc&es pour que de nombreusCs editions de ce livre se repandent
rapidement dans le public mathematique et parmi la jeunesse
studieuse de nos university.”
[1016]( 2 )
Fnrtn li. A. Roberts, M.A., T.C.D.
“I am much obliged for tie copy of your book on * Conics.’
I have been much interested in the many new examples it con-
tains. I hope you will republish some time your memoir on
‘ Bicircular Quartics,’ with Additions.”
From Edward J. Routh, M.A., F.R.S., &c., Cambridge.
“ It seems to me that so excellent a treatise will soon make its
way to the front, even against the severe competition which all
books on Conics now meet with. I fancy that many of the
theorems cannot be elsewhere found so conveniently explained.’*
From M. Emile Lemoine, Paris.
“ Je m’empresse de vous remercier de votre amiable envoi, et
de vous exprimer combien je suis flatte d’etre cite par un maitre
dans un ouvrage classique.”
From M. Maurice D’Ocagne, Paris.
“ Je m’empresse de vous faire parvenir mes remerciments pour
l’honneur que vous me faites en me citant dans votre remarquable
ouvrage.”
From the Freeman’s Journal.
“ This treatise exhibits in a marked degree the qualities which
distinguish the author’s other works. It is at once compact and
comprehensive. .	.	. Dr. Casey’s treatise, indeed, may well
accomplish for this generation what Dr. Salmon’s did for their
fathers, namely, to introduce the young mathematician to the
latest developments in the highest departments of the Science.
.	.	. Scarcely any important step is there in the work which
he has not simplified, giving one, and sometimes several, original
methods. .	.	. The method of projection is treated of in an
entirely original manner, not requiring the consideration of space
of three dimensions, the projection being performed in one plane,
and the equation of the curve obtained by a simple transformation.
The properties of the M‘Cay, Neuberg, and Brocard Circles are
considered at some length: indeed the latter has now a literature
of its own, to which Dr. Casey has largely contributed. . . ·
There are considerably over a thousand examples, graduated
from easy applications of the formulae to exercises of the highest
class of difficulty. Many of these are original, and many are
historically interesting. ’ ’
From the Dublin Evening Mail.
“ This work will give a new impulse to the study of Analytic
Geometry, introduce the student to new and more powerful
methods, and greatly enlarge his mathematical horizon. It has
been known for some time that Dr. Casey was engaged on a
‘ Conic Sections,’ and people expected that notwithstanding the
many works in the field, Dr. Casey’s would present a good many
valuable novelties; but the work has, we venture to think, ex-( 3 )
celled all anticipation. .	.	. Dr. Casey makes a large use of
determinants. He introduces them in the very first chapter, and
employs them to the end. In the first chapter we find a section
devoted to complex variables, and Gauss’s geometrical represen-
tation of them ; and from Clebsch he takes the comparison of
* point ’ and ‘ line ’ co-ordinates. He gives a great development
to trilinear co-ordinates and tangential equations, discusses the
new circles (Tucker’s, Brocard’s, Neuberg’s, M‘Cay’s), applies
Aronhold’s notation to the discussion of the general equation of
the second degree, and reproduces from his own Papers in the
Transactions of learned societies many important theorems and
problems relating to inscribed and circumscribed figures, ortho-
gonal conics, and the tact-invariant of two-conics. .	.	. The
work is a noble monument to Dr. Casey’s genius, and his master-
ing of all the resources of Modem Analytic Geometry.”
From Nature.
“ Dr. Casey, by the publication of this third treatise, has
quite fulfilled the expectations we had formed when we stated
some months since that he was engaged upon its compilation.
It is a worthy companion of those which have preceded it. It
possesses many points of novelty, i.e. for the English mathe-
matician. He has from the first introduction of certain recent
•continental discoveries in geometry taken a warm interest in
them, and in the purely geometrical treatment of them, has him-
self given several beautiful proofs, and has added discoveries of
his own. We may here note that this work has met with a very
warm welcome in France and Belgium. The author himself has
■added so much in years now long past to several branches of the
subject treated of in the volume under notice—the equation of
the circle (and of the conic) touching three circles (three conics),
and other properties—that he is specially fitted, by his intimate
acquaintance with it, and by his long tuitional experience, to
write a hook on analytical geometry.
“ The divisions are into eight chapters, the first of which, in
four sections, treats of the point, three sections being taken up
with Cartesian and polar co-ordinates, and the transformation of
•co-ordinates; the fourth section gives a brief account of complex
variables introduced by Cauchy in 1825.	‘ The introduction of
these variables is one of the greatest strides ever made in mathe-
matics.’ The second chapter on the Right Line treats it (§ 1) by
Cartesian, then (§ 2) by trilinear, and (§ 3) by point and line co-
ordinates ; this last comparison is taken from Clebsch’s * Vorle-
sungen fiber Geometrie.’ In Chapter III. four sections are
devoted to the circle, § 2 being devoted to a system of tangential
circles; §,3 to the ‘ trilinear’ forms of equations to the old circles
and to all the recent circles; § 4 is devoted to tangential equations.
Chapters iv., v., vi., vn. treat of, respectively, the general
equation of the second degree, the parabola, ellipse, and hyper-
bola ; Chapter viii. (Miscellaneous investigations) discusses
many matters of novelty and interest ; § 1 is on contact of( 4 )
conic sections; § 2, similar figures, gives a good resume of result»
connected with Brocard’s points and circles, Neuberg’s circles,
M‘Cay’s circles, and Kiepert’s hyperbola ; in § 3 on the general
equation interlinear co-ordinates. Aronhold’s notation*is ‘now
published for the first time in an English treatise on Conic Sec-
tions.* The remaining six sections are occupied respectively
with Envelopes, Projections, Sections of a Cone, Homographic
Divisions, Reciprocal Polars, and Invariants and Covariants. An
idea has now, we trust, been conveyed to the reader of the
ground covered by Dr. Casey; a good deal of it is, of course,
well-worn ground, but even this has been adorned by his touch,
and much relating to the new circles has never before been intro-
duced into our books: these circles must soon become as familiar
to our junior students as the nine-points circle, whose properties
are by this time nearly exhausted.
“ The examples are exceedingly numerous, and a good feature·
is that most of the results obtained in them are numbered con-
secutively with the important results of the text. This enables
the author to refer to them with facility.’*
From the Educational Times, May, 1886.
“ The eminence and experience of the author are a sufficent
guarantee for careful work and original research, and in this
volume of moderate size there is a great deal that will be fresh to
English readers . . .
“ There can be no doubt that this treatise ought to be studied
as a whole by everyone to acquaint himself with the most recent
developments of Algebraical Geometry.*’
From the Educational Times, September, 1886.
“In this book the author has added to those propositions
usually met with in Treatises on Analytical Geometry many
which we have seen in no other books on the subject; notably
extending the equations of circles inscribed in and circumscribed
about triangles to polygons of any number of sides, and extend-
ing to Conics the properties of circles cutting orthogonally. The
demonstrations are concise and neat. In many cases the author
has substituted original methods of proof advantageously, and in
some has also added the old methods. We would specially note
his treatment of the General Equation of the Second Degree,
which is more satisfactory than many we have seen. Throughout
the book there are numerous exercises on the subject matter, and
at the end of each section a collection of problems bearing on
that part of the subject. These problems have been obtained
from Examination Papers and other sources. They have been
selected with much care and judgment. The name of the pro-
poser has in many cases been added, and will cause more interest
to be taken in the solution.”