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# Elementary triangle goemetry

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### Elementary triangle goemetry

1. 1. 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. 2. 2
3. 3. 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 Tan ( 1 A ) , tan ( 1 B ) and tan ( 1 C ) 2 2 2 in terms of a , b and c. Cos ( 1 A ) , cos ( 1 B ) and cos ( 1 C ) in terms of a , b and c. 2 2 2 Sin ( 1 A ) , sin ( 1 B ) and sin ( 1 C ) in terms of a , b and c. 2 2 2 §8: Further Triangle Relationships. 37 Relationship between r and R. Relationships between rA , rB , rC and R. The distances AI , BI and CI . The distances AI A , BI B and CI C . The distances II A , II B and II C . §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. 4. 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. 5. 5 Motivating Problem 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. B A typical diagram is shown in Figure MP.1 a h c P C A x b-x Figure MP.1 The area of the triangle, ∆ , is seen to be ∆ = 1 ( base ) ( ⊥ height ) 2 (MP.1) ∆ = 1 bh, 2 where b = AC and h = 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: PB sin C = BC h sin C = a Therefore, h = a sin C (MP.2) From (MP.1) and (MP.2) we have the general area formula ∆ = 1 ab sin C 2 (MP.3)
6. 6. 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. a 2 = x 2 + h2 (MP.4) c 2 = ( b − x ) + h2 . 2 and (MP.5) Eliminating h from (MP.4) and (MP.5) gives: a2 − x2 = c2 − (b − x ) 2 a2 − c2 = x2 − (b − x ) . 2 That is: Thus, by the difference of two squares formula we have a 2 − c 2 = ( x − b − x )( x + b − x ) a 2 − c 2 = ( x − b + x )( x + b − x ) a 2 − c 2 = ( 2 x − b )( b ) Hence, a2 + b2 − c2 x= . (MP.6) 2b Substituting (MP.6) into (MP.4) gives 2  a2 + b2 − c2  h = a − 2 2  .  2b  That is 4 a 2b 2 − ( a 2 + b 2 − c 2 ) 2 h = 2 (MP.7) 4b 2 Once again, by the difference of two squares formula we have the alternative form of (MP.7): h 2 = ( 2ab − a 2 )( + b2 − c 2 2ab + a 2 + b 2 − c 2 ), 4b 2 = ( 2ab − a 2 − b 2 + c 2 )( 2ab + a 2 + b2 − c 2 ) , 4b 2 = (c 2 − a 2 + 2ab − b 2 )( a 2 + 2ab + b2 − c 2 ) , 4b 2
7. 7. 7 That is: h 2 = (c 2 − a 2 − 2ab + b2 )( a 2 + 2ab + b 2 − c 2 ), 4b 2 which, on factorising gives: h 2 = (c 2 − (a − b) 2 ) (( a + b) 2 − c2 ). 4b 2 Therefore, on using the difference of two squares formula again we have: h2 = ( c − a − b )( c + a − b )( a + b − c )( a + b + c ) 4b 2 ( c − a + b )( c + a − b )( a + b − c )( a + b + c ) h2 = . (MP.8) 4b 2 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 ( a + b + c ) ! 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: ( c − a + b ) ≡ ( a + b + c − 2a ) ( c + a − b ) ≡ ( a + b + c − 2b ) (MP.9) ( a + b − c ) ≡ ( a + b + c − 2c ) Realising that ( a + b + c ) is just the perimeter of the original triangle ABC, say p gives (MP.8) as: ( p − 2a )( p − 2b )( p − 2c )( p ) h2 = . (MP.10) 4b 2 However, if we then let the new variable, s, be defined as the semi-perimeter then (MP.10) is re-written ( 2s − 2a )( 2s − 2b )( 2s − 2c )( 2s ) h2 = . 4b 2
8. 8. 8 This is then easily factorised to give: 16 ( s − a )( s − b )( s − c )( s ) h2 = 4b 2 or 4s ( s − a )( s − b )( s − c ) h2 = . b2 2 s ( s − a )( s − b )( s − c ) Hence, h= (MP.11) b Finally then, substituting (MP.11) back into (MP.1) gives  2 s ( s − a )( s − b )( s − c )  ∆ = 1 b×  2  b    or ∆ = s ( s − a )( s − b )( s − c ) (MP.12) Which is the familiar result of Heron of Alexandria (First Century A.D)
9. 9. 9 “What a marvel that so simple a figure as the triangle is so inexhaustible in its properties!” (A. L. Crelle, 1821)
10. 10. 10
11. 11. 11 §1: Basic Trigonometrical Formulae sin θ = tan θ (1.1) cos θ sin 2 θ + cos 2 θ = 1 (1.2) θ 0 30 45 60 90 1 2 3 sin θ 0 1 2 2 2 (1.3) 3 2 1 cos θ 1 0 2 2 2 1 tan θ 0 1 3 ∞ 3 1 cosec θ = sin θ 1 sec θ = (1.4) cos θ 1 cot θ = tan θ tan 2 θ + 1 = sec 2 θ (1.5) 1 + cot 2 θ = cosec 2 θ (1.6)
12. 12. 12
13. 13. 13 §2: Further Trigonometrical Formulae sin ( A + B ) = sin A cos B + cos A sin B (2.1) sin ( A − B ) = sin A cos B − cos A sin B (2.2) cos ( A + B ) = cos A cos B − sin A sin B (2.3) cos ( A − B ) = cos A cos B + sin A sin B (2.4) tan A + tan B tan ( A + B ) = (2.5) 1 − tan A tan B tan A − tan B tan ( A − B ) = (2.6) 1 + tan A tan B P+Q   P−Q  sin P + sin Q = 2 sin    cos   (2.7)  2   2  P +Q   P −Q  sin P − sin Q = 2 cos    sin   (2.8)  2   2  P+Q  P−Q  cos P + cos Q = 2 cos    cos   (2.9)  2   2  P +Q   P −Q  cos P − cos Q = − 2 sin    sin   (2.10)  2   2  sin ( 2 A ) = 2sin A cos A (2.11) cos ( 2 A ) = cos2 A − sin 2 A = 2cos2 A − 1 = 1 − 2sin 2 A (2.12) sin 2 A = 1 2 (1 − cos ( 2 A) ) (2.13) cos 2 A = 1 2 (1 + cos ( 2 A) ) (2.14) Notice further, if we define t = tan 1 θ then it can be shown that 2 2t 1 − t2 2t sin θ = , cosθ = and tan θ = . (2.15) 1 + t2 1+ t 2 1 − t2
14. 14. 14
15. 15. 15 §3: Ratio Theorems Theorem (3.1) p a λ p + µ q λ a + µb If we have that = then = , for any numbers λ , µ , m and n. q b mp + nq ma + nb Proof: p a Let = ≡t q b ⇒ p = qt , a = bt . λ p + µq λ ( qt ) + µ q λt + µ Therefore, = = mp + nq m ( qt ) + nq mt + n and λ a + µb λ ( bt ) + µ q λt + µ = = □ ma + nb m ( bt ) + nq mt + n Theorem (3.2) p a λ p + µa p a If we have that = then = = , for any numbers λ and µ . q b λ q + µb q b Proof: p a Let = ≡t q b ⇒ p = qt , a = bt . λ p + µ a λ ( qt ) + µ ( bt ) t ( λ q + µ b ) p a Therefore, = = =t ≡ ≡ □ λ q + µb λ q + µb ( λ q + µb ) q b
16. 16. 16
17. 17. 17 B §4: Basic Triangle Formulae Cosine Rule c h a A C b x y Figure 4.1 x y Notice that: cos A = , cos C = c a Therefore, x = c cos A, y = a cos C Hence b = x + y ≡ c cos A + a cos C Using symmetry we interchange the variables to yield the complete set of results thus: a a = b cos C + c cos B b = c cos A + a cos C (4.1) c b c = a cos B + b cos A The formulae of (4.1) are known as the Projection Formulae. If we now multiply the equations of (4.1) by a, b and c, respectively, we have: a 2 = ab cos C + ac cos B (4.2) b2 = bc cos A + ab cos C (4.3) c 2 = ac cos B + bc cos A (4.4) Now construct (4.3) + (4.4) - (4.2) to give: b2 + c 2 − a 2 = 2bc cos A Therefore, a 2 = b2 + c 2 − 2bc cos A (4.5) Equation (4.5) is known as the Cosine Rule for triangles. Symmetry yields the other forms: b2 = c 2 + a 2 − 2ca cos B and c 2 = a 2 + b2 − 2ab cos C
18. 18. 18 Sine Rule Also from Figure 4.1 we have the further set of relations: h h sin A = , sin C = c a Therefore, h = c sin A or h = a sin C Hence, c sin A = a sin C ≡ h sin A sin C Therefore, = a c However, the initial orientation of the triangle ABC was arbitrary sin A sin B sin C ⇒ = = (4.6) a b c Equation (4.6) is known as the Sine Rule for triangles. The Sine Rule can be extended by considering a circle through the apexes of the triangle ABC (known as the Circumcircle of the triangle ABC) B B P A C O O A C P Figure 4.2 Figure 4.3 In both Figure 4.2 and 4.3 the red lines AP and PC have been added to the original Circumcircle problem. In both cases the line segment AP is draw so as to pass through the centre of the Circumcircle and is therefore a diameter. ⇒ ACP is a Right-Angle in both figures (Angle in a Semi-Circle is a right- angle). Further, APC = ABC since angles subtended by a single chord in the same segment of a circle are equal (Euclid Book III Prop. 21). AC b Therefore, from Figure 4.2 we have: sin ( APC ) ≡ sin ( B ) = ≡ AP 2 R where R is the radius of the Circumcircle of triangle ABC.
19. 19. 19 From Figure 4.3 APC = 180 − B (Cyclic Quadrilateral) Therefore, sin ( APC ) = sin (180 − B ) ≡ sin B Hence, as for Figure 4.2 we have b sin B 1 sin B = or = 2R b 2R Therefore, from (4.6) we have: sin A sin B sin C 1 = = = (4.7) a b c 2R Tangent Rule b sin B From (4.7) we have that = c sin C Using Theorem (3.1) with λ ≡ m ≡ n = 1 and µ = −1 this relation can be written (1) b + ( −1) c (1) sin B + ( −1) sin C = (1) b + (1) c (1) sin B + (1) sin C b − c sin B − sin C That is = b + c sin B + sin C Using (2.8) and (2.7) this can be rewritten as B +C   B −C  2 cos    sin   b−c  2   2 , = b + c 2 sin  B + C  cos  B − C       2   2  B −C  sin    b−c  2 × 1 = , b + c cos  B − C  sin  B + C       2   2  B+C  cos     2  B−C  tan    b−c  2 . = b + c tan  B + C     2 
20. 20. 20 b−c That is tan 1 ( B − C ) = tan 1 ( B + C ) (4.8) b+c 2 2 However, A + B + C = 180 ⇒ 1 2 ( B + C ) = 90 − 1 A 2 (4.9) From (4.9) we see that tan 1 ( B + C ) = tan ( 90 − 1 A ) 2 2 tan(90 ) − tan 1 A = 2 1 + tan(90 ) tan 1 A 2 tan 2 A 1 1− tan(90 ) = 1 + tan 1 A 2 tan(90 ) 1− 0 = 0 + tan 1 A 2 ∴ tan 1 ( B + C ) = cot 1 A 2 2 Hence (4.8) can be written in its alternative form b−c tan 1 ( B − C ) = cot 1 A (4.10) b+c 2 2 Equation (4.10) is known as the Tangent Rule for triangles. Symmetry yields the other forms: c−a a tan 1 ( C − A ) = cot 1 B c+a 2 2 c b a−b tan 1 ( A − B ) = cot 1 C a+b 2 2
21. 21. 21 §5: Other Triangle Formulae B Area Formulae in terms of a, b and c. c h a A C b Figure 5.1 The area of triangle ABC is found from h ∆ = 1 bh, where sin C = 2 , a = 1 b ( a sin C ) 2 Hence, ∆ = 1 ab sin C 2 (5.1) a By symmetry, ∆ = 1 ab sin C = 1 bc sin A = 1 ca sin B 2 2 2 c b c From (4.7) we have that sin C = 2R c  Therefore, (5.1) becomes ∆ = 1 ab  2    2R  abc That is: ∆= (5.2) 4R abc or R= . 4∆ a sin B Also, from (4.7) we have that b= sin A Therefore, (5.1) now becomes: a sin B  ∆ = 1 a 2   sin C  sin A  sin B sin C That is: ∆ = 1 a2 2 sin A Hence, a sin B sin C 1 2 sin C sin A 1 2 sin A sin B ∆ = 1 a2 2 = 2b = 2c (5.3) c b sin A sin B sin C
22. 22. 22 Sin A in terms of a, b and c, etc. From (1.2) sin 2 A = 1 − cos 2 A = (1) − ( cos A ) 2 2 = (1 − cos A )(1 + cos A ) Therefore, from (4.5) this becomes  b2 + c 2 − a 2   b2 + c 2 − a 2  sin 2 A =  1 −  1 +   2bc  2bc      2bc − b 2 + c 2 − a 2   2bc + b 2 + c 2 − a 2  =    2bc  2bc      2bc − b 2 − c 2 + a 2   2bc + b 2 + c 2 − a 2  =    2bc  2bc   a − b + 2bc − c   b + 2bc + c − a  2 2 2 2 2 2 =    2bc  2bc   a − (b − c )   (b + c ) − a  2 2 2 2 =    2bc  2bc     1 ( )( = 2 2 a2 − (b − c ) (b + c ) − a2 4b c 2 2 ) 1 = 2 2 ( a − b + c )( a + b − c )( b + c − a )( b + c + a ) 4b c 1 = 2 2 ( a + b + c − 2b )( a + b + c − 2c )( b + c + a − 2a )( b + c + a ) 4b c Now let s = 1 ( a + b + c ) , the Semi-perimeter, then we have 2 1 sin 2 A = ( 2s − 2b )( 2 s − 2c )( 2s − 2a )( 2s ) 4b 2 c 2 1 = 2 2 16 ( s − b )( s − c )( s − a )( s ) 4b c Hence 4 sin 2 A = s ( s − a )( s − b )( s − c ) bc 2 2 Therefore, 2 a sin A = s ( s − a )( s − b )( s − c ) bc 2 sin B = s ( s − a )( s − b )( s − c ) (5.4) c b ca 2 sin C = s ( s − a )( s − b )( s − c ) ab
23. 23. 23 Notice further that these three identities from (5.4) could be written 2a sin A = s ( s − a )( s − b )( s − c ) ≡ 2aK abc 2b sin B = s ( s − a )( s − b )( s − c ) ≡ 2bK (5.5) abc 2c sin C = s ( s − a )( s − b )( s − c ) ≡ 2cK abc From which, the Sine Rule can be deduced, since sin A sin B sin C ≡ 2K  ≡ 1  = =   a b c  2R  Moreover, from (5.1) and (5.4) we have that 2 ∆ = 1 ab × sin C ≡ 1 ab × 2 2 s ( s − a )( s − b )( s − c ) ab Hence ∆ = s ( s − a )( s − b )( s − c ) (5.6) This is the triangle area formula met previously in (MP.7): Heron’s Formula. We can now use the notation of (5.6) or more simply the form of (5.1) to write: Sin A, sin B, sin C in terms of a, b and c. a 2∆ 2∆ 2∆ sin A = , sin B = , sin C = (5.7) c b bc ca ab Cos A, cos B, cos C in terms of a, b and c. From (4.5) we simply rearrange to yield a b2 + c2 − a 2 c2 + a 2 − b2 a 2 + b2 − c2 cos A = , cos B = , cos C = (5.8) c b 2bc 2ca 2ab Tan A, tan B, tan C in terms of a, b and c. From (1.1), (5.5) and (5.8) we have a 4∆ 4∆ 4∆ tan A = , tan B = 2 , tan C = 2 (5.9) c b b +c −a 2 2 2 c +a −b 2 2 a + b2 − c2
24. 24. 24
25. 25. 25 §6: Associated Circles A Incircle Let I be the Incentre of the F E triangle ABC, obtained by bisecting the interior angles of the triangle ABC. Then I ID = IE = IF ≡ r , where r is the radius of the incircle. ( ID, IE and IF are the ⊥ 's . C B D from I to the respective sides) Figure 6.1 We have that Area of triangle ABC = Areas of triangles ( BIC + CIA + AIB ) ∆ = 1 ar + 1 br + 1 cr 2 2 2 ∆ = 1 r (a + b + c) 2 Hence for s, the semi perimeter ∆ = rs (6.1) ∆ or r= (6.2) s Now, let AE = AF ≡ x  Tangents to a circle from  BD = BF ≡ y  a single point have equal  CD = CE = z  lengths ⇒ x+ y+ y+z+z+x= p ( perimeter ) Hence x+ y+z =s ( semi perimeter ) But BD + DC = y + z ≡ a ⇒ x+a=s Hence x = s−a. Therefore, we can show: a x=s−a y = s−b (6.3) c b z = s−c
26. 26. 26 A From triangle ABC we see that s-a s-a IF r tan ( 1 A ) = = FA s − a 2 F Similarly E r a tan ( 1 A ) = s-b s-c s−a 2 I r tan ( 1 B ) = (6.4) s−b 2 c b D C r tan ( 1 C ) = s-c s−c B s-b 2 Figure 6.2 Moreover, from triangles AIE, BIF and CID we have AE = AF ≡ s − a = r cot ( 1 A ) a 2 BD = BF ≡ s − b = r cot ( 1 B ) 2 (6.5) c b CD = CE ≡ s − c = r cot ( 1 C ) 2 Therefore, as a = BD + DC then a = r cot ( 1 B ) + r cot ( 1 C ) 2 2 = r ( cot ( 1 B ) + cot ( 1 C ) ) 2 2  cos ( 1 B ) cos ( 1 C )  = r  sin ( 1 B ) + sin ( 1 C )  2 2   2 2   cos ( 1 B ) sin ( 1 C ) + sin ( 1 B ) cos ( 1 C )  = r  2 2 2 2    sin ( 1 B ) sin ( 1 C ) 2 2  sin ( 1 B + 1 C ) =r 2 2 sin ( 2 B ) sin ( 1 C ) 1 2 But A + B + C = 180 and therefore, 1 2 B + 1 C = 90 − 1 A , giving 2 2 sin ( 90 − 1 A ) a=r 2 sin ( 1 B ) sin ( 1 C ) 2 2 That is cos ( 1 A ) a=r 2 . sin ( 1 B ) sin ( 1 C ) 2 2 Hence sin ( 1 B ) sin ( 1 C ) r=a 2 2 (6.6) cos ( 1 A ) 2
27. 27. 27 Together, we have sin ( 1 B ) sin ( 1 C ) r=a 2 2 a cos ( 1 A ) 2 sin ( 1 C ) sin ( 1 A ) r =b 2 2 (6.7) c b cos ( 1 B ) 2 sin ( 1 A ) sin ( 1 B ) r =c 2 2 cos ( 1 C ) 2 From (4.7) we have sin A 1 = , etc, where R is the Circumcircle. a 2R or a = 2 R sin A , etc Thus (6.6) becomes sin ( 1 B ) sin ( 1 C ) r = 2 R sin A ⋅ 2 2 cos ( 1 A ) 2 Using (2.11) gives sin ( 1 B ) sin ( 1 C ) r = 2 R × 2sin ( A ) cos ( A ) ⋅ 1 1 2 2 cos ( 1 A ) 2 2 2 Hence r = 4 R sin ( 1 A ) sin ( 1 B ) sin ( 1 C ) 2 2 2 (6.8)
28. 28. 28 Excircles A Let I A be the centre of a circle, opposite angle A, of radius rA obtained by bisecting the angles B and C externally and A internally. I We have that: B D A' D' C Area of triangle ABC = E' Areas of triangles ( AI A B + CI A A − BI AC ) That is IA ∆ = 1 crA + 1 brA − 1 arA 2 2 2 F' = 1 rA ( c + b − a ) 2 = 1 rA ( a + b + c − 2a ) 2 Figure 6.3 Hence for s, the semi perimeter ∆ = rA ( s − a ) (6.9) Similarly, we have ∆ rA = a s−a ∆ rB = (6.10) c b s−b ∆ rC = s−c Notice further that A, I and I A are collinear as both AI and AI A lie on the bisector line of angle A Now let AE ' = AF ' ≡ x A  Tangents to a circle from  BD ' = BF ' ≡ y A  a single point have equal  CD ' = CE ' = z A  lengths Then AF '+ AE ' = AB + BF '+ AC + CE ' 2 xA = c + y A + b + z A ⇒ = b + c + ( yA + z A ) but y A + z A = BD '+ D ' C ≡ BC = a .
29. 29. 29 Therefore: 2 xA = a + b + c xA = 1 ( a + b + c ) 2 Hence xA = s ( semi perimeter ) (6.11) A In a similar manner a xB = s and xC = s c b The three distances x A = s, xB = s and xC = s are displayed below D A' B D' C E' A IA F' IC B C Figure 6.4 Figure 6.5 IB A Figure 6.6 B C
30. 30. 30 Notice also from triangle AI A F ' and from (6.11) that I A F ' rA tan ( 1 A ) = 2 = AF ' s ⇒ rA = s tan ( 1 A ) . 2 Similarly a rA = s tan ( 1 A ) 2 rB = s tan ( 1 B ) 2 (6.12) c b rC = s tan ( 1 C ) 2 Therefore, we can further show that a AE ' = AF ' ≡ s = rA cot ( 1 A ) , 2 s = rB cot ( 1 B ) , 2 (6.13) s = rC cot ( 1 C ) . c b 2 Moreover, since BF ' = BD ' = AF '− AB = s − c and CE ' = CD ' = AE '− AC = s − b then DD ' = CD ' ∼ CD = ( s − b) ∼ ( s − c) =b∼c Where D is the point of tangency of the incircle with side a, D ' is the point of tangency of the excircle with side a, and where the symbol ∼ is the positive difference between its two arguments. Namely: DD ' = b ∼ c ≡ b − c (6.14) Further, as BD ' = CD = s − c it follows that A ' is the midpoint of DD ' where A ' is also the midpoint of BC.
31. 31. 31 Heron’s Area Formula IA C D I D' A B Figure 6.7 As BI and BI A bisect the angle B both internally and externally, it follows that IBI A = 90 . Moreover, since IDB and I A D ' B are also right angles then this implies that triangles BID and BI A D ' are similar. Namely: Triangles BID ≅ I A BD ' BI ID BD Thus = = I A B BD ' I A D ' BI r s −b that is: = = I AB s − c rA Therefore: r ⋅ rA = ( s − b )( s − c ) (6.15) On using (6.2) and (6.10) this becomes ∆ ∆ ⋅ = ( s − b )( s − c ) s s−a Hence ∆ = s ( s − a )( s − b )( s − c ) (6.16) Which is the form of the area of triangle ABC as met previously in (MP.7) and (5.6)
32. 32. 32
33. 33. 33 §7: Further Triangle Formulae Y IA C I A D B X Figure 7.1 Tan ( 1 A ) , tan ( 1 B ) and tan ( 1 C ) in terms of a , b and c. 2 2 2 ID r From Figure 7.1 we have tan ( 1 A ) = ≡ AD s − a 2 On using (6.15) this becomes  ( s − b )( s − c )    tan ( 1 A ) =  rA  = ( s − b )( s − c ) s−a rA ( s − a ) 2 On using (6.9) we have ( s − b )( s − c ) ( s − b )( s − c ) tan ( 1 A ) = = 2 ∆ ∆ ( s − a) (s − a) ( s − b )( s − c ) ( s − b )( s − c ) Hence: tan ( 1 A ) = = s ( s − a )( s − b )( s − c ) s (s − a) 2 Similarly, ( s − b )( s − c ) tan ( 1 A ) = s(s − a) 2 a ( s − c )( s − a ) tan ( 1 B ) = (7.1) s ( s − b) 2 c b ( s − a )( s − b ) tan ( 1 C ) = s (s − c) 2
34. 34. 34 Cos ( 1 A ) , cos ( 1 B ) and cos ( 1 C ) in terms of a , b and c. 2 2 2 From Figure 7.1 we have I ACY = 1 2 (180 − C) = 90 − 1 C 2 = 1 ( A + B) 2 Also, using the construction properties of the Incircle and Excircle I A AC = 1 A 2 Hence AI AC = 180 − ( 1 A + (180 − 2 I ACY ) ) = 180 − ( 1 A + (180 − 1 ( A + B ) ) ) 2 2 = 180 − (180 − 1 B ) 2 AI AC = 1 B ≡ 2 DBI Therefore, triangles AIB and ACI A are similar, that is AIB ≅ ACI A . AI IB AB = = AC CI A AI A ⇒ AI IB c = = b CI A AI A Hence AI × AI A = bc (7.2) AX Also from Figure 7.1 cos ( 1 A ) = 2 (7.3) AI A AD and cos ( 1 A ) = 2 (7.4) AI The product of (7.3) and (7.4) gives AX AD cos ( 1 A ) × cos ( 1 A ) = 2 2 × AI A AI That is AX × AD cos2 ( 1 A ) = AI A × AI 2 On using (6.3), (6.11) and (7.2) this becomes s × (s − a) cos 2 ( 1 A ) = 2 bc Hence s(s − a) cos ( 1 A ) = 2 a bc s ( s − b) cos ( 1 B ) = 2 (7.5) c b ca s (s − c) cos ( 1 C ) = 2 ab
35. 35. 35 Sin ( 1 A ) , sin ( 1 B ) and sin ( 1 C ) in terms of a, b and c. 2 2 2 Using (1.1), (7.1) and (7.5) we have that sin ( 1 A ) = tan ( 1 A ) cos ( 1 A ) 2 2 2 Hence ( s − b )( s − c ) s(s − a) sin ( 1 A ) = × s(s − a) 2 bc Giving ( s − b )( s − c ) sin ( 1 A ) = 2 a bc ( s − c )( s − a ) sin ( 1 B ) = 2 (7.6) c b ca ( s − a )( s − b ) sin ( 1 C ) = 2 ab Notice that in each of the above three subsections the negative root is rejected if the angles, A, B and C are those of a triangle.
36. 36. 36
37. 37. 37 §8: Further Triangle Relationships Relationship between r and R A Note that BIC = 180 − 1 B − 1 C 2 2 I = 180 − 1 ( B + C ) 2 r = 180 − 1 (180 − A ) 2 B C D = 90 + A1 2 Figure 8.1 ID r and that sin ( 1 B ) = 2 ≡ IB IB Hence, r = IB sin ( 1 B ) 2 (8.1) Using the Sine Rule of (4.6) we have that sin ( 1 C ) sin ( 90 + 1 A ) sin ( 1 B ) = = 2 2 2 , IB BC IC and from the first two of these fractions we have sin ( 1 C ) cos ( 1 A ) 2 = 2 . IB BC BC sin ( 1 C ) Therefore: IB = 2 , cos ( 1 A ) 2 a sin ( 1 C ) that is IB = 2 . (8.2) cos ( 1 A ) 2 Substituting (8.2) into (8.1) gives  a sin ( 1 C )  r = cos ( 1 A )  sin ( 2 B ) 2 1   2  a sin ( 2 C ) sin ( 1 B ) 1 = 2 cos ( 2 A ) 1 2 R sin A sin ( 1 B ) sin ( 1 C ) Using (4.7) a can be written as: a = 2 R sin A, giving r = 2 2 . cos ( 1 A ) 2 Using (2.11) gives 2 R × 2sin ( 1 A ) cos ( 1 A ) sin ( 1 B ) sin ( 1 C ) r= 2 2 2 2 . cos ( 1 A ) 2 Hence: r = 4 R sin ( 1 A ) sin ( 1 B ) sin ( 1 C ) 2 2 2 (8.3) as met previously in (6.8).
38. 38. 38 Relationships between rA , rB , rC and R. From Figure 8.2 we see that C I AD ' r sin ( I A BD ') = ≡ A I AB I AB D' IA rA that is rA = I A B sin ( 90 − 1 B ) 2 (8.4) rA = I A B cos ( 1 B ) 2 A B Figure 8.2 Using the Sine Rule of (4.6) we have that sin ( 90 − 1 C ) sin ( 90 − 1 A ) sin ( 90 − 1 B ) = = 2 2 2 I AB BC I AC and from the first two of these fractions we have cos ( 1 C ) cos ( 1 A ) 2 = 2 I AB BC BC cos ( 1 C ) Therefore, I AB = 2 . cos ( 1 A ) 2 a cos ( 1 C ) that is, I AB = 2 (8.5) cos ( 1 A ) 2 Substituting (8.5) into (8.4) gives  a cos ( 1 C )   cos ( 1 A )  cos ( 2 B ) rA =  2 1   2  a cos ( 1 C ) cos ( 1 B ) = 2 2 cos ( 2 A ) 1 Using (4.7) a can be written as: a = 2 R sin A , 2 R sin A cos ( 1 C ) cos ( 1 B ) giving rA = 2 2 . cos ( 1 A ) 2 Using (2.11) gives 2 R × 2sin ( 1 A ) cos ( 1 A ) cos ( 1 C ) cos ( 1 B ) rA = 2 2 2 2 . cos ( 1 A ) 2 Hence: a rA = 4 R sin ( 1 A ) cos ( 1 B ) cos ( 1 C ) , 2 2 2 and similarly rB = 4 R cos ( 1 A ) sin ( 1 B ) cos ( 1 C ) , 2 2 2 (8.6) c b rC = 4 R cos ( 1 A ) cos ( 1 B ) sin ( 1 C ) . 2 2 2
39. 39. 39 The distances AI , BI and CI . From Figure 8.3 notice that C ID IA sin ( 1 A ) = 2 I IA A D B X Therefore, Figure 8.3 a AI = r cosec ( 1 A ) 2 BI = r cosec ( 1 B ) 2 (8.7) c b CI = r cosec ( 1 C ) 2 If we just consider the triangle AI A X for the present and determine some of its angles from Figure 8.4 IA we find that: Clearly AIB = 180 − 1 ( A + B ) 2 I and so on using the Sine Rule of (4.6) we have from triangle AIB A D B X sin ( B ) sin ( AIB ) sin ( A ) 1 1 2 = = 2 Figure 8.4 AI AB BI Considering just the first two fractions here gives AB sin ( 1 B ) AI = 2 sin ( AIB ) Therefore c sin ( 1 B ) AI = 2 sin (180 − 1 ( A + B ) ) 2 That is on using (2.2): c sin ( 1 B ) AI = 2 sin ( 1 ( A + B ) ) 2 c sin ( 1 B ) = 2 sin ( 90 − 1 C ) 2 c sin ( 1 B ) = 2 cos ( 1 C ) 2 But on using (4.7) c can be written as: c = 2 R sin C , giving
40. 40. 40 2 R sin C sin ( 1 B ) AI = 2 cos ( 1 C ) 2 2 R × 2sin ( 1 C ) cos ( 1 C ) sin ( 1 B ) Using (2.11) gives AI = 2 2 2 cos ( 1 C ) 2 Hence AI = 4 R sin ( 1 B ) sin ( 1 C ) a 2 2 BI = 4 R sin ( 1 C ) sin ( 1 A ) 2 2 (8.8) c b CI = 4 R sin ( 1 A ) sin ( 1 B ) 2 2 which give an alternative form of (8.7). Further, from (8.7), as it is somewhat r easier we may proceed AI = r cosec ( 1 A ) = sin ( 1 A ) 2 2 Thus, from (7.6) we have r r bc AI = = ( s − b )( s − c ) ( s − b )( s − c ) bc r s ( s − a ) bc ≡ s ( s − a )( s − b )( s − c ) r s ( s − a ) bc s ( s − a ) bc Using (5.6) and (6.1) AI = ≡ . ∆ s ( s − a ) bc (s − a) Hence, AI = = bc a s s a ( s − b) c ( s − b) similarly BI = = ca (8.9) c b s s ab ( s − c ) (s − c) CI = = ab s s Note that the relations of (8.9) can be written abc ( s − a ) (s − a) AI = ≡K , s a a abc ( s − b ) ( s − b) abc BI = ≡K , where K = (8.10) s b b s abc ( s − c ) (s − c) CI = ≡K . s c c (s − a) ( s − b) (s − c) ⇒ AI : BI : CI = : : (8.11) a b c
41. 41. 41 The distances AI A , BI B and CI C . C IA I A D B X Figure 8.5 From Figure 8.5 notice that I AD sin ( 1 A ) = 2 IAA Therefore, AI A = rA cosec ( 1 A ) a 2 BI B = rB cosec ( 1 B ) 2 (8.12) c b CI C = rC cosec ( 1 C ) 2 C Notice also from Figure 8.6 that IA θ + θ + φ + φ = 180 θ + φ = 90 I Hence IBI A = 90 Also, since D B θ = 1 B and φ = 1 (180 − B ) = 90 − 1 B 2 2 2 Figure 8.6 then I I A B = 90 − 1 ( A + B ) 2 IA I Figure 8.7 A D B
42. 42. 42 On using the Sine Rule of (4.6) we have from triangle ABI A sin ( 90 + 1 B ) sin ( AI A B ) sin ( 1 A ) = = 2 2 AI A AB BI A Considering just the first two fractions here gives AB sin ( 90 + 1 B ) AI A = 2 sin ( AI A B ) Therefore c cos ( 1 B ) AI A = 2 sin ( 90 − 1 ( A + B ) ) 2 That is on using (2.2): c cos ( 1 B ) AI A = 2 cos ( 1 ( A + B ) ) 2 c cos ( 1 B ) = 2 cos ( 90 − 1 C ) 2 c cos ( 1 B ) = 2 sin ( 1 C ) 2 But on using (4.7) c can be written as: c = 2 R sin C giving 2 R sin C cos ( 1 B ) AI A = 2 sin ( 2 C ) 1 Using (2.11) gives 2 R × 2sin ( 1 C ) cos ( 1 C ) cos ( 1 B ) AI A = 2 2 2 sin ( 1 C ) 2 Hence a AI A = 4 R cos ( 1 B ) cos ( 1 C ) 2 2 and similarly BI B = 4 R cos ( 1 C ) cos ( 1 A ) 2 2 (8.13) c b CI C = 4 R cos ( 1 A ) cos ( 1 B ) 2 2 which give an alternative form of (8.12) Further, from (8.12), as it is slightly easier we may proceed rA AI A = rA cosec ( 1 A ) = sin ( 1 A ) 2 2
43. 43. 43 Thus, from (6.10) and (7.6) we have ∆ 1 AI A = ( s − a ) ( s − b )( s − c ) bc s ( s − a )( s − b )( s − c ) bc = (s − a) ( s − b )( s − c ) bc s ( s − a ) = (s − a) bc s = (s − a) bc = (s − a) s Therefore: bc or AI A = (s − a) s a ca similarly BI B = (8.14) c b ( s − b) s ab CI C = (s − c) s In a manner similar to the processes used to arrive at (8.11) we can also show that 1 1 1 AI A : BI B : CI C = : : (8.15) a(s − a) b( s − b) c(s − c) Further, if we take respective products of (8.9) and (8.14) then we quickly get the very nice results: a AI × AI A = bc BI × BI B = ca (8.16) c b CI × CI C = ab
44. 44. 44 The distances II A , II B and II C . IA I Y A D B X Figure 8.8 Consider the similar triangles AID and II AY , we have AI AD ID = = II A DX I AY That is AI (s − a) r = = II A AX − AD I A X − YX AI (s − a) r = = II A s − ( s − a ) rA − ID AI ( s − a ) r Hence, = = (8.17) II A a rA − r From the first two fractions we have a  II A =    AI a s−a b  similarly II B =    BI (8.18) c b  s−b c  II C =    CI s−c Therefore, on using the values from the first forms in (8.10), the results in (8.18) can be written in the manner a  abc ( s − a ) II A =   × s−a s a abc a II A = s (s − a) Hence abc a abc b abc c II A = , II B = , II C = . (8.19) s (s − a) s ( s − b) s (s − c)
45. 45. 45 Squaring the value of II A from (8.19) gives  abc   a  ( II A ) 2 =    s  (s − a)  Therefore, from the last two fractions of (8.17) this becomes abc  rA − r  ( II A ) = 2    (8.20)  s  r  abc From (5.2) we have ∆= 4R and from (6.1) that ∆ = rs abc Hence, 4 Rr = (8.21) s Therefore, from (8.20) and (8.21) we have that ( II A ) = 4 R ( rA − r ) , ( II B ) = 4 R ( rB − r ) , ( IIC ) = 4 R ( rC − r ) 2 2 2 (8.22) Moreover, using the half-angle formula in (7.5) we see that (8.19) can be rearranged thus: a 2bc bc 1 a II A = =a =a ≡ s(s − a) s(s − a) s ( s − a ) cos ( 1 A ) 2 bc Hence a b c II A = , II B = , II C = . (8.23) cos ( A ) 1 2 cos ( B ) 1 2 cos ( 1 C ) 2 Therefore, on using the results of (4.7), (8.23) can be rewritten in a further, alternative form II A = 4 R sin ( 1 A ) , 2 II B = 4 R sin ( 1 B ) , 2 II C = 4 R sin ( 1 C ) . 2 (8.24)
46. 46. 46
47. 47. 47 §9: Further Triangle Centres The Orthocentre of any Triangle ABC The perpendiculars drawn from the vertices of a triangle ABC to the opposite sides are concurrent at a point called the Orthocentre, H. A From Figure 9.1 let AD, BE and CF be the perpendiculars F on BC, CA and AB respectively, and H the Orthocentre; then DH = BD tan ( HBD ) E H = AB cos B × tan ( 90 − C ) , = c cos B cot C Using (4.7) c can be written as: B D C Figure 9.1 c = 2 R sin C , giving DH = 2 R sin C cos B cot C Hence a DH = 2 R cos B cos C hA EH = 2 R cos C cos A hB (9.1) c b FH = 2 R cos A cos B hC DA = AB sin B Further, = c sin B = 2 R sin C sin B That is a DA = 2 R sin B sin C EB = 2 R sin C sin A (9.2) c b FC = 2 R sin A sin B Therefore, from (9.2) and (9.1) we have that HA = DA − DH = 2 R sin B sin C − 2 R cos B cos C Using (2.3) this becomes HA = −2 R cos ( B + C ) = −2 R cos (180 − A ) Hence: a HA = 2 R cos A HB = 2 R cos B (9.3) c b HC = 2 R cos C
48. 48. 48 Notice that HA2 = ( 2 R cos A ) 2 = 4 R 2 cos 2 A = 4 R 2 (1 − sin 2 A ) = 4 R 2 − 4 R 2 sin 2 A = 4 R 2 − ( 2 R sin A ) 2 Using (4.7) a can be written as: a = 2 R sin A , giving HA2 = 4 R 2 − a 2 Hence, we have the alternative forms of (9.3) as a HA = 4 R 2 − a 2 HB = 4 R 2 − b2 (9.4) c b HC = 4 R 2 − c 2 Exercise: Establish (9.1) to (9.4) for an obtuse-angled triangle Again using Figure 9.1, note the following simply found forms of hA , hB and hC . Area of triangle ABC = 1 BC × AD 2 = 1 a × hA 2 a Therefore, 2∆ 2∆ 2∆ hA = , and similarly hB = , hC = . a b c c b