HEMATICAL CES LIBRARY. UNIVERSITY OF CALIFORNIA AT LOS ANGELES 685 Heath's Mathematical Monographs Issu.-d under the general ^ .-.'. Webster Wells, S. B Professor of Mathematics ill Che Massachuse i: . ' : - Some Noteworthy : roperties oi the Triangu a nd its Circle* V. f i. BKUX'l . VI . I ii - - Of MAIHEMA; -. '-' STATK N< >i HALS D. C. Heath & Co., Publisher,- Boston New York Chicago Number 8 ; Price, Ten Cents Th nner s . ' :-.:-. N > a MPmHl^^SSffl^^^^TO^^B^^B ' ' & $9jlniJiflfNm ^^^^^^^B K Clot) d teachi :ructure oi p.l to becorru little ; ving fultoanv :'UN LIICAGO HEATH'S MATHEMATICAL MONOGRAPHS NUMBER 8 SOME NOTEWORTHY PROPERTIES OF THE TRIANGLE AND ITS CIRCLES BY W. H. BRUCE, A.M., PH.D. PROFESSOR OF MATHEMATICS, NORTH TEXAS STATE NORMAL SCHOOL DENTON, TEXAS BOSTON, U.S.A. D. C. HEATH & CO., PUBLISHERS 1902 COPYRIGHT, 1902, BY D. C. HEATH & Co. O /^ "-3'neering & Mathematica/ V X Sciences ^ THE TRIANGLE 1. The triangle is the only rigid figure ; that is to say, it is the only polygon whose sides alone determine its shape and size, or whose angles determine the ratio of its sides and of all other parts. 2. Every triangle has three sides, three angles, three altitudes, three medians, three interior angle bisectors, three exterior angle bisectors, and three * perpendicular bisectors of its sides. In addition * to these parts of every triangle, there are certain , other lines and circles dependent upon these parts ; and their mutual relations and the variety of their ' interdependencies are fascinating to the student just awakening to the beauties of the science of geometry. 3. In general, a triangle is determined by any three of its parts, one of these parts being a length, whether this length be a side, an altitude, the radius of any one of its numerous dependent circles, or any other length. 3 4 The Triangle 4. One of the simplest of the many interesting relations between the various parts of the triangle is that of the concurrence of certain corresponding lines. 5. The perpendicular bisectors of the sides of a triangle are concurrent in a point. This point, being the centre of the circumscribed circle, is called the circumcentre of the triangle. From the definition, it is evident that the circumcentre of an acute triangle is within the triangle ; that of a right triangle is the mid-point of the hypotenuse; and that of an obtuse triangle is without the triangle. 6. The altitudes of a triangle are concurrent in a point called the orthocentre of the triangle. Because the perpendicular is the shortest sect from a point to a given line, the orthocentre of an acute triangle is within the triangle ; that of a right triangle at the vertex of the right angle ; and that of an obtuse triangle without the triangle. 7. The medians of a triangle are concurrent in that trisection point of each which is the second from a vertex. This point of concurrence is called the centroid of the triangle, and it is evident that the point is always within the triangle. 8. The bisectors of the interior angles of a tri- angle are concurrent. This point of concurrence and its Circles is the centre of the inscribed circle, and is there- fore called the in-centre of the triangle. It is evi- dent that the in-centre is always within the triangle. 9. The bisectors of the interior and the ex- terior angles of a triangle are concurrent four times by threes. \v Proof. Let ABC (Fig. i) be any A, and let the bisectors of interior angles B and C meet at O. Then since BO is the locus of points equidistant from AB and BC, and CO the locus of points equidistant from BC and AC, O is equidistant from the three sides. Similarly let bisectors of exterior angles B and C meet at O', then O' is equidistant from AB, BC, and AC. Similarly for O" and O'". 10. Hence O', O", O'", are centres of circles touching the sides of triangle ABC. These circles 6 The Triangle are called the escribed circles of the triangle, and O', O", O'", the ex-centres of the triangle. 11. The extremities of one side of a triangle and the feet of the altitudes to the other two sides are concyclic (i.e. they lie on the same circle). Fig. a. Proof. Let ABC (Fig. 2) be any A, and AD, BE, CP y its altitudes. Then A, B, D, E t are con- cyclic, also B, C, E, F, and A, C, D, F. Proof. The circle on AB as diameter passes through the vertices of the right angles D and E. Therefore A, B, D, E, are concyclic. Similarly for B, C, E, F, and for A, C, D, F. 12. The vertex of any angle of a triangle, the feet of the altitudes to its adjacent sides, and the orthocentre are concyclic. Proof. The circle on BK (Fig. 3) as diameter passes through vertices of right angles D and F. Therefore B, D, K, F, are concyclic. Similarly for C, D, K, E, and for A, E, K, F. Fig. 3. 13. Any two altitudes of a triangle cut each other so that the product of the segments of the one is equal to the product of the segments of the other. Proof. By 11, A, B, D, -E(Fig. 2), are concyclic. .-. AKxKD = BKx KE. (If two chords of a O intersect, the product of the segments of the one is equal to the product of the segments of the other.) COLLINEARITY. 14. Definition. Collinear points are those that lie on the same straight line. 15. The circumcentre, the centroid, and the orthocentre of a triangle are collinear, and the sect between the former two is half the sect between the latter two. Proof. Let P be the circumcentre and K the orthocentre of A ABC (Fig. 4). Draw MN and PK, and also median AN cutting PK at O. The Triangle Fig. 4. A MPN is similar to A AKC (sides of one being parallel respectively to sides of the other), (M and N being mid-points of sides AB and BC). ANPO and AAOK are similar, and NP = \AK, hence, ON= \AO, or AO = \ AN. .-. By 7, O is centroid of A ABC. Hence, the circumcentre, P, the centroid, O, and the orthocentre, K, are collinear, and PO = % KO. 16. Corollary. Since in similar triangles NPM and AKC(Fig. 4), NM=\AC, PM=\CK, we may say, The perpendicular sect from the circumcentre to a side of the triangle is equal to half the sect from the opposite vertex to the orthocentre. 17. The circle through the mid-points of the sides of a triangle passes also through the feet and its Circles 9 of the altitudes, and bisects the sects between the vertices and the orthocentre. Let L, My TV (Fig. 4) be the mid-points of the sides, D, E, .Fthe feet of the altitudes, of AAJ3C, and S, T, Z the mid-points of CK, AK, and B K, respectively; then are the nine points, Z, M, N\ D, E, F; S, T, Z; concyclic. Proof. Draw NT\ APGWand TGK are similar. (Sides of one being parallel respectively to sides of the other.) Also, by 16, NP = \AK= TK; .: GK = GPy and GT= GN. Then since G is the mid-point of PK, and T the mid-point of AK, GT=^AP, radius of circum- scribed O. Hence, with G as a centre and radius \AP, a O will pass through 71 Similarly, it is proved to pass through 5 and Z. But GN= GT, therefore, this circle passes through Ny and its diameter is TN. Similarly, it is proved to pass through L and M. Again, this circle on 77V, the hypotenuse of rt. A TDN, as diameter, passes through D. Similarly, it is proved to pass through E and F. .'. L, M, N\ D, E, F-, S, Ty Z\ are concyclic. 18. The circle through the nine-points, men- tioned in 17, is called the nine-point circle of the triangle. 10 The Triangle 19. Corollary. The centre of the nine-point circle of a triangle is collinear with the circumcentre, the centroid, and the orthocentre, and is midway between the circumcentre and the orthocentre. 20. Corollary. The radius of the nine-point circle of a triangle is half the radius of the cir- cumscribed circle. 21. The nine-point circle of a triangle bisects every sect between its orthocentre and its circum- scribed circle. C~ Fig- 5- Let P (Fig. 5) be the circumcentre, and K the orthocentre of A ABC, and HS a chord of the circumscribed O through K, cutting the nine-point circle at X and V. Then V is the mid-point of KS, and X the mid-point of KH. and its Circles n Proof. From G, the centre of the nine-point O, draw GX and G V. Draw also PH and PS. InAPKS, GV=^PS(2Q). InAPKH, GX = \PH=\PS = GV. Therefore, no other sect can be drawn from G to HS which will be equal to GV or \PS. But a sect from G to mid-point of KS = \ PS, there- fore, V is the mid-point of KS, and X the mid- point of KH. 22. The sect from the centroid of a triangle to the circumscribed circle is twice the prolongation of that sect from the centroid to the nine-point circle. Let RL (Fig. 5) be a chord of the circumscribed circle through Q, the centroid of A ABC, cutting the nine-point circle at E and T. To prove QR = 2 QT and QL = 2 QE. Proof. By 15, PQ = \PK. By 19, PG = \ PK. Hence, PQ-\ PG, or PQ = 2 GQ. In &PQRand GQT, also PQ = 2GQ; therefore, a sect from G, parallel to PR and cutting QL is equal to J PR, 12 The Triangle But, by 20, GT=\PR and GE = \PR, there- fore no other sect from G to RL is equal to \ PR. (From a given point only two equal sects can be drawn to a given line.) .-. GT is the parallel to PR. Hence, &PQR and GQT are similar, and con- sequently QR = 2QT. Similarly, QL = 2 QE. 23. The sect from the vertex of any angle of a triangle to the point in which an altitude to one of its including sides intersects the circumscribed circle is equal to the sect from this vertex to the orthocentre. Fig. 6. Let K be the orthocentre of &ABC, and let AK prolonged cut the circumscribed O at P, then will CP=CK. Proof. By 17, the nine-point circle of &ABC passes through D. By 21, KD = DP. and its Circles 13 .-. CD is the perpendicular bisector of KP. Hence, CK=CP. 24. The three circles, each of which passes through two vertices and the orthocentre of a triangle, are each equal to the circumscribed circle. G Fig. 7. Let A D, BE, CF (Fig. 7), be the altitudes and K the orthocentre of A ABC. Prolong CF to meet the circumscribed circle at G. To prove the circumscribed circle of A.ABK is equal to the circumscribed circle of A ABC. Proof. Draw AG and B G. KG is perpendicular to AB by hypothesis. KF=FG by 21 (F being on the nine-point O). .', A K = A G and BK = BG. Hence, &AKB = &ABG. (Having three sides of one = three sides of other.) The Triangle /. The circumscribed O of A ,4^ = the cir- cumscribed O of AAJ3G. But the circumscribed O of A ABC is also that of A ABG. .'. The circumscribed O of AAKB is equal to circumscribed O of A ABC. Similarly for circumscribed of &AKC and BKC. 25. Each centre of the inscribed or an escribed circle of a triangle is the orthocentre of the tri- angle having the other three centres as vertices. Let O be the in-centre, O', O", O'" the ex-centres (Fig. i) of &ABC. To prove O is orthocentre of A6>'<9"<9'", O' the orthocentre of AOO"O'", O" the orthocentre of AOO'O'", and O'" the orthocentre of A OO'O". and its Circles 15 Proof. y.BAO"'=y. CA O" (each being half of equal angles). (each being half ZBAC). Hence, O'A is perpendicular to O"O'". Similarly, O"B and O'"C are altitudes of AO'O"O'". /. O is the orthocentre of A O'O"O'". Similarly, O"C, O'"B, and OA are altitudes of AOO"O'", therefore O', their point of concur- rence, is the orthocentre of AOO"O'". And similarly, O" is the orthocentre of &OO'O'", and O'" is the orthocentre of A OO'O". 26. The four circles each of which passes through three of the centres of the inscribed and escribed circles of a triangle are equal. Let O, O', O", O'" (Fig. i) be the in-centre and the ex-centres of &ABC. Then, by 24 and 25, the circle through O'OO'" is equal to the circle through O'O"O'", and simi- larly for the circles through O'OO" and O"OO'" respectively. 27. The circumscribed circle of a triangle is the nine-point circle of each of the four triangles formed by joining the in-centre and the ex-centres of the triangle. 1 6 The Triangle Let O (Fig. i) be in-centre, and O r , O", O'" the ex-centres, of A ABC. By 25, O'A isO"O'"; O"BO'O'"; O'"C. O'O". .'. A, B, and C are the feet of the altitudes of A OO'O", OO'O'", OO"O'", and O'O"O"'. Hence, the circumscribed O of A ABC passes through the feet of the altitudes of the four triangles. .'. By 17-18, the circumscribed O of A ABC is the nine-point circle of the four triangles desig- nated. 28. The circumscribed circle of a triangle bisects the sects between the in-centre and each ex-centre, and also the sect between any two ex-centres of the triangle. Let ABC (Fig. i) be a triangle, O its in-centre, and O', O", and O'" its ex-centres. To prove the circumscribed circle of A ABC bisects OO', OO", OO'", O'O", O'O 1 ", and O"O'". Proof. By 27, the circumscribed O of A ABC is the nine-point circle of each of the A OO'O", OO'O'", OO"O"', and O'O"O'". By 25, O, O', O", and O'" are each the ortho- centre of the A having the other three as vertices. Therefore, by 21, this nine-point circle bisects OO', OO", OO'", O'O", O'O'", and O"O'". and its Circles 17 29. The nine-point circle has become celebrated in the history of geometry. Its discovery is usu- ally accredited to Euler (1707-1783). It was first called "the nine-point circle" in 1842 by M. Terquem, editor of Nouvelles Annales, this name being derived from its property of hav- ing as nine points of its circumference the feet of the three altitudes, the mid-points of the three sides, and the mid-points of the three sects from the orthocentre to the vertices. In 1822 Professor K. W. Feuerbach of Nurem- berg published his discovery and demonstration of the remarkable property of the nine-point circle, now known as the " Feuerbach Theorem." 30. The Feuerbach Theorem. The nine-point circle of a triangle is tangent to the in-circle and each of the ex-circles of the triangle. 31. The nine-point circle of a triangle is tangent to each of the sixteen inscribed and escribed circles of the four triangles formed by joining the in- centre and the ex-centres of the original triangle. Proof. By 27, the circumscribed O of a A is the nine-point O of the four A formed by joining its in-centre and ex-centres. By the Feuerbach Theorem, the nine-point O is tangent to each of the inscribed and escribed CD. i8 The Triangle 32. The feet of the perpendiculars from any point on a circle to the sides of an inscribed triangle are collinear. Fig. 8. Let ABC (Fig. 8) be an inscribed A, P any point on the circumscribed O, and PD, PE, /'/''perpen- diculars to the sides of the A. To prove that D t E, and Fare collinear. Proof. Draw PA, PB t PC, EF, and ED. In quadrilateral PECF, since %E and ^-F are both right A, the quadrilateral is cyclic (see any text), and, similarly, PADE is cyclic (i.e. its vertices are concyclic). (being inscribed A on the same arc of O PECF) . (each being supplement of % PCB). and its Circles 19 Hence, PEF+ % FED = PAB + But /M + />.> = st (being opposite ^ of quadrilateral inscribed in O FADE). .-. % PEF+ % PED = st. . Hence DEF is a straight line. 33. Definition. The line joining the feet of the perpendiculars from a point on a circle to the sides of an inscribed triangle is called the pedal line of the point and the triangle. 34. The pedal line of a point and a triangle bisects the sect between such point and the ortho- centre of the triangle. Let PD, PE, and PF (Fig. 9) be J to sides of &ABC inscribed in a circle through P and let O be the orthocentre of A ABC, then will DF bisect OP. Proof. Prolong altitude CG to meet the O at H. Draw AP. Draw PH cutting AB at M and DF at N. Draw OM. Since AFP and AEP are both rt by hy- pothesis, the G on AP as diameter passes through E and F. .-. PF.E =%PAC=% PHC= % FPN (PF and CH being II). Hence A F/W is isosceles, or PN = 20 The Triangle H ^*" Fig. g. In A OMG and HMG, OG= GH (as G is on nine-point circle and H on circumscribed circle of &ASQ. .-. OM= MH and OMG = HMG. But y.HMG is complement of ~%.MHG and I MFN is complement of FPN, which equals .: % HMG and its equals, ^ OMG and each = X ^fRV. /. F^V= MN. Hence, P^= J/7V. Also, since Of = ^ NFM, DFis parallel to OM. .-. Since DF bisects PJ/it also bisects OP. and its Circles 21 35. The sect joining the orthocentre of an in- scribed triangle and a point on the circumscribed circle and the pedal line of the point and the circle intersect each other on the nine-point circle of the triangle. Proof. By 34, DF(F\g. 9) bisects OP. By 21, the nine-point O of A ABC bisects OP. .'. the sect OP, the pedal line DF, and the nine- point O all have a common point S. 36. The sect joining the orthocentres of two triangles that have the same base and equal ver- tical angles is parallel and equal to the sect join- ing the vertices of the vertical angles. Fig. 10. In &ABC and DEC (Fig. 10), let 4 A = % >, K the orthocentre of A ABC, and .S the ortho- centre of A DEC. Then will KS be || and = AD. 22 The Triangle Proof. Since the A have the same base and equal vertical angles, they have the same circumscribed O. Let P be circumcentre of the triangles, and draw PMLBC. By 16, AK and DS each = 2. PM. .'. AK = DS t hence, AKSD is a parallelogram. .-. KS || AD and = AD. 37. If ABCD is a cyclic quadrilateral, the sects joining the orthocentres of the four triangles ABC, ABD, ACD, and BCD form a quadrilateral congruent with ABCD. Fig. 10. Proof. Let ABCD (Fig. 10) be a cyclic quad- rilateral ; K, S, orthocentres of A ABC and BDC. By 36, KS = and || AD. Similarly, the sects joining any other two ortho- centres = and || to a side of ABCD ; therefore, the and its Circles 23 quadrilateral whose vertices are the orthocentres of the four triangles designated is equal to or con- gruent with ABCD. 38. If ABCD is a cyclical quadrilateral, the sects joining the centres of the nine-point circles of the triangles ABC, ABD, ACD, and BCD form a quadrilateral similar to ABCD and equal to one- fourth ABCD. Proof. In Fig. 10, draw PK and PS. By 19, the centre of the nine-point O of A ABC is the mid- point of PK, and the centre of the nine-point circle of &.BCD is the mid-point of PS. The sect joining the mid-points of PK and PS is parallel and equal to \ KS, and therefore parallel and equal to \AD. Similarly, the sects joining the centres of the other nine-point circles are respectively parallel and equal to half a side of ABCD. .'. the quadrilateral thus formed is similar to and equals \ABCD. 39. If A, B, C, D are coney clic, the pedal lines of each point and the triangle whose vertices are the other three, also the sects joining each of these points to the orthocentre of the triangle whose vertices are the other three, and the nine-point circles of the four triangles, all pass through a common point. 24 The Triangle Proof. By 36, AKSD (Fig. 10) is a parallelo- gram. /. AS and DK bisect each other. Hence, the sect joining any one of the four points to the ortho- centre of its corresponding triangle bisects AS and DK; therefore they all pass through H, the mid-point of AS. By 21, the nine-point circle of each triangle passes through the mid-point of the sect from its orthocentre to the circumscribed circle. Hence, the nine-point circles of the four triangles all pass through H. By 34, the pedal lines of each point and the triangle whose vertices are the other three points all pass through H, the mid-point of AS. FORMULAS OF THE TRIANGLE 40. Let a, b, c, denote the sides of a triangle; 2 s, its perimeter; A, its area; //, h n ', h'", the altitudes ; m' , m" , m"', its medians ; /, /", /"', its angle bisectors ; /', /", /'", the perpendiculars for the circumcentre to the sides ; R, the radius of the circumscribed circle ; r, the radius of the inscribed circle ; r 1 , r", r'", the radii of the escribed circles. 2 41. K = - ~vs(s a) (s ) (s c). a 42. A = V-r<> - ) - b) - 0- and its Circles ak* bk" cti" 43. A = = -- = -- 22 2 > b + c 44. 45. 46. * = 4A 47. r = - -\/bcs(s-a). 48. r " = __ j a J b s c (For proofs of 43-50, see any text.) 49. In an acute triangle, / +/' +/" = R + r. Proof. C Let ABC (Fig. 11) be a A, Z>, E, F mid-points of its sides, P the circumcentre. Denote BC by a, AC by b, AB by c, PE by /, PF by /', by p'. 26 The Triangle BDPE is a cyclic quadrilateral. Then BE x PD + BD x PE = DE x (The product of the diagonals of an inscribed quadrilateral is equal to the sum of the product of its opposite sides.) Or, or, ap'" + cp' = Z>R. (i) Similarly, ap" + bp' = cR, (2) and bp'" + cp"=aR. (3) Again, ap' + bp" + cp" = 2 A = ar + br + cr. (4) Adding and factoring, (a + + ^)(/ +/' +/") = ( + b + ;)(tf + r). Reducing, p' +p" +/" =R + r. 50. In case of an obtuse triangle, p 1 " being drawn to longest side, / +p" p"' = R + r. 51 I + i+-L = l. W A * . | It* III yi yii Y n " + f" Substituting for b and ^ their values h'fi" h'p'" ,, or, P + -jrr + it!*-= h ' P' P" P'" Hence, F + F" + F 7 " = 55. rr' + fr" + rr>" + r'r" + r'r"' + r"r"' ab-\-ac-\- be. 28 The Triangle Substituting for r, r', r", r'", their values in 49 and 48, the first member of the equation becomes A 2 A 2 A 2 A 2 s(s a) s(s b) s(s c) (s a)(s b) A 2 A 2 (s a)(s c) (s &)(s c) Substituting value of A in 42, and reducing, = (s- b}(s -c) + (s- a)(s - c) + (s - a)(s - b) + s(s c) + s(s b} + s(s a) = ( s - *)[(* - <0 + - ay] + (s - a}[(s - c} + s] + sl(s-c} + (s-b}-] = (s-b%2s-a- c*)+(s - ay(2s- c}+s(2s-b-c} = bs b 2 + as + bs a 2 ab + as = ab + ac + be 56. A=V?rW". DESCRIPTIVE PRICE LIST OF WELLS' NEWEST BOOKS BOUND IN HALF LEATHER. LARGE PAGE. CLEAR TYPE. ACADEMIC ARITHMETIC 348 pages. Price, $1.00. ESSENTIALS OF ALGEBRA With or without answers. 3,300 problems. 375 pages. Price, #1.10. NEW HIGHER ALGEBRA With or without answers. 460 pages. Price, $1.32. 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