elementary_triangle_goemetry1.pdf

Elementary Triangle Geometry Mark Dabbs The Mathematical Association Conference University of York, U.K Spring 2004 Version 1.1 April 2004 (www.mfdabbs.com) 2 3 Contents Motivating Problem. 5 §1: Basic Trigonometrical Formulae. 11 §2: Further Trigonometrical Formulae. 13 §3: Ratio Theorems. 15 Theorem (3.1). Theorem (3.2). §4: Basic Triangle Formulae. 17 Cosine Rule. Sine Rule and Cicumcircle. Tangent Rule. §5: Other Triangle Formulae. 21 Area Formulae in terms of a, b and c. Sin A , sin B and sin C in terms of a, b and c. Cos A , cos B and cos C in terms of a, b and c. Tan A , tan B and tan C in terms of a, b and c. §6: Associated Circles. 25 Incircle. Excircles. Heron’s Area Formula. §7: Further Triangle Formulae. 33 () () ( ) 11 1 22 2 Tan , tan and tan in terms of , and .AB C abc () () ( ) 11 1 22 2 Cos ,cos and cos in terms of , and .AB C abc () () ( ) 11 1 22 2 Sin ,sin and sin in terms of , and .AB C abc §8: Further Triangle Relationships. 37 Relationship between and .Rr Relationships between , , and . ABC rrr R The distances A , and C .IBI I The distances A , and C . AB C IBI I The distances , and . AB C II II II §9: Further Triangle Centres. 47 The Orthocentre of any Triangle ABC . The Pedal Triangle of any Triangle ABC . The Circumcircle and the Pedal Triangle. The Excentric Triangle. 4 §10: Special Cevian Lengths. 55 The Centroid and Medians of any Triangle. Cevians Bisecting Angles Internally. Cevians Bisecting Angles Externally. Apollonius’ Theorem. A Generalisation of Apollonius’ Theorem – Stewart’s Theorem §11: Problems. 63 Appendix: Concurrences of Straight Lines in a Triangle. 67 Circumcentre. Incentre. Centroid. Orthocentre. Bibliography. 73 5 Motivating Problem a c h x b - x P C B A The motivation for this work came from an open question to a class to find the area of a triangle whose base is known but whose perpendicular height is not known. A typical diagram is shown in Figure MP.1 Figure MP.1 The area of the triangle, , is seen to be ∆ ( ) ( ) 1 2 1 2 base height ,bh ∆= ⊥ ∆= (MP.1) where and h . bAC= BP= After some discussion, two methods were proposed o Method 1: Using Trigonometry o Method 2: Using Pythagoras Method 1 is perhaps the more familiar and progresses thus: In triangle BPC we have: sin sin PB C B C h C a = = Therefore, sinha C = (MP.2) From (MP.1) and (MP.2) we have the general area formula 1 2 sinab C ∆ = (MP.3) 6 Method 2 was somewhat more involved and led to quite a voyage of discovery! Note the following Pythagorean relations within the two triangles CBP and ABP . 22 axh 2 = + (MP.4) and () 2 2 cbxh 2 = −+. (MP.5) Eliminating h from (MP.4) and (MP.5) gives: () 2 222 axc bx−=−− That is: . () 2 22 2 ac x bx−=−− Thus, by the difference of two squares formula we have ( ) ( ) ()( ()() 22 22 22 2 ac xbxxbx ac xbxxbx ac xbb −=−− +− −=−+ +− −= − ) Hence, 22 2 abc x b 2 + − = . (MP.6) Substituting (MP.6) into (MP.4) gives 2 222 22 2 abc ha b  +− =−   . That is ( ) 2 22 2 2 2 2 2 4 4 ab a b c h b −+− = (MP.7) Once again, by the difference of two squares formula we have the alternative form of (MP.7): () ( ) 222 222 2 2 22 4 ab a b c ab a b c h b −+− ++− = , () ( ) 222 222 2 22 4 ab a b c ab a b c b −−+ ++− = , () ( ) 22 22 22 2 22 4 c a ab b a ab b c b −+ − + +− = , 7 That is: () ( ) 22 22 22 2 2 22 4 c a ab b a ab b c h b −− + + +− = , which, on factorising gives: () () () ( ) 22 22 2 2 4 cab abc h b −− + − = . Therefore, on using the difference of two squares formula again we have: () ( ) ( ) ( ) 2 2 4 cabcababcabc h b −− +− +− ++ = () ( ) ( ) ( ) 2 2 4 cabcababcabc h b −+ +− +− ++ = . (MP.8) Now it’s time to ask which of the four factors in the numerator “looks” the “nicest” and hope that the answer to come back is the fourth or last one of () ! abc++ Having established this, the suggestion is then made that it is a pity that the other three factors do not have this same elegant symmetry and once agreed that we ought to insist that such symmetry exist in these other three factors. It is eventually determined that a suitable “trick” is to rewrite them in the following manner: () ( ) ()( ()( 2 2 2 cab abc a cab abc b abc abc c −+ ≡ ++− +− ≡ ++− +− ≡ ++− ) ) ) (MP.9) Realising that ( abc + + is just the perimeter of the original triangle ABC , say p gives (MP.8) as: ( ) ( ) ( ) ( ) 2 2 222 4 p ap bp cp h b − −− = . (MP.10) However, if we then let the new variable, s , be defined as the semi-perimeter then (MP.10) is re-written () ( ) ( ) ( ) 2 2 2222222 4 sasbscs h b − −− = . 8 This is then easily factorised to give: ( ) ( ) ( ) ( ) 2 2 16 4 sasbscs h b − −− = or ( ) ( ) ( ) 2 2 4ss a s b s c h b − −− = . Hence, ()()() 2 ss a s b s c h b − −− = (MP.11) Finally then, substituting (MP.11) back into (MP.1) gives ()()() 1 2 2 ss a s b s c b b  −−− ∆= ×    or (MP.12) ()()() ss a s b s c∆= − − − Which is the familiar result of Heron of Alexandria (First Century A.D) 9 “What a marvel that so simple a figure as the triangle is so inexhaustible in its properties!” (A. L. Crelle, 1821) 10 . II §9: Further Triangle Centres. 47 The Orthocentre of any Triangle ABC . The Pedal Triangle of any Triangle ABC . The Circumcircle and the Pedal Triangle. The Excentric Triangle. . 15 Theorem (3.1). Theorem (3.2). §4: Basic Triangle Formulae. 17 Cosine Rule. Sine Rule and Cicumcircle. Tangent Rule. §5: Other Triangle Formulae. 21 Area Formulae in terms of. Elementary Triangle Geometry Mark Dabbs The Mathematical Association Conference University of