Modern geometry . B X O fig. 15. AIBC+ AICA+ AIAB = = AABC Now AIBC =i^ix. BC AICA = \rb, AIAB = \ro; .•. |m + ^rb + lrc-- = A; , a + b ti-A .*. rs = A J Ar = —. 26 the tuiangleTheorem fig. 16. A 1;CA + A IjAB - A liBC = A ^ow A liCA = ^IjYi. CA A liAB = |riC,A liBC = ^r^a; h + o — a 2 = A. But 6 + c — a = (a + 6 + c)-2ffl= 2s - 2a;.*. ri (s - a) = A;A Similarly r, =-^^, r,= ?^^. THE TRIANGLE 27 1 SR Ex. 48. Prove that in an equilateral triangle r=^R, r\=ri=ri=-^. Ex. 48. If the ex-centres be joined, the triangle so formed is similar tothe triangle XYZ. Ex. SO. Prove that the circle
Modern geometry . B X O fig. 15. AIBC+ AICA+ AIAB = = AABC Now AIBC =i^ix. BC AICA = \rb, AIAB = \ro; .•. |m + ^rb + lrc-- = A; , a + b ti-A .*. rs = A J Ar = —. 26 the tuiangleTheorem fig. 16. A 1;CA + A IjAB - A liBC = A ^ow A liCA = ^IjYi. CA A liAB = |riC,A liBC = ^r^a; h + o — a 2 = A. But 6 + c — a = (a + 6 + c)-2ffl= 2s - 2a;.*. ri (s - a) = A;A Similarly r, =-^^, r,= ?^^. THE TRIANGLE 27 1 SR Ex. 48. Prove that in an equilateral triangle r=^R, r\=ri=ri=-^. Ex. 48. If the ex-centres be joined, the triangle so formed is similar tothe triangle XYZ. Ex. SO. Prove that the circle on llj as diameter passes through B and construct a triangle, having given BC, z B, and the length iii, Ex. 51. AZi + AYi = AB-|-AC + BC. Ex. 62. AYi = AZi = s. AZ + AY = AB + AC-BO. AY = AZ=s-a. Ex. 55. ZZi = YYi=a. Ex. 56. BXi = CX=s-c. BX = CX, = s-6. Ex. 68. XXx=o ~ 6. Theorem 13. (i) AYi = AZi = s. (ii) AY = AZ = s - a. (iii) YYi = ZZi = a. (iv) BXj = CX = s-o. (v) XXi = c~6. 28
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