ags@pucc-h (Dave Seaman) (11/17/85)
I posted a proof for the geometry problem which depended on a formula for the length of a bisector as a function of the sides. This is the derivation of that formula. Let a triangle have sides a, b and c. Let t be the length of the bisector which meets side c. Let s = (a + b + c) / 2. Let u be either of the two small angles formed by the angle bisector. The triangle with sides a, b and c has an angle 2u between sides a and b. Its area is therefore A = (1/2) ab sin 2u. The two small triangles formed by the bisector have areas A1 = (1/2) at sin u and A2 = (1/2) bt sin u where t is the length of the bisector. Since A = A1 + A2, we have (1/2) ab sin 2u = (1/2) at sin u + (1/2) bt sin u or ab sin u cos u = (1/2) (a+b) t sin u which yields 2ab cos u t = ---------. (1) a + b Applying the law of cosines to the large triangle: 2 2 2 2 a +b -c 2 cos u - 1 = cos 2u = -------- 2ab 2 2 2 2 a +2ab+b -c 2 cos u = ------------ 2ab 2 2 2 (a+b) -c cos u = --------- 4ab a+b+c a+b-c 1 = ----- * ----- * -- 2 2 ab s(s-c) = ------ ab Substituting in (1): 2 t = --- ab cos u a+b 2 / s(s-c) \ = --- ab sqrt | ------ | a+b \ ab / 2 = --- sqrt(abs(s-c)) a+b This was the formula used in the proof. -- Dave Seaman ..!pur-ee!pucc-h!ags