Triangle ex-radius properties
Let's say a triangle ABC has 3 ex-circles. One of them is touching side BC at D and the extended line segments of AB and AC at E,F. Then Important Ex Radius Result Number 1: BD = s - c, CD = s - b and AE = AF = s. Where AB = c, BC = a and AC = b. Proof: And s = semi perimeter = (a + b + c)/2 Why? By equal tangents property: AE = AF, BE = BD and CD = CF Let BE = x => BD = x and CD = a - x = CF AE = c + x AF = b + a - x AE = AF => c + x = b + a - x => x = (b + a - c)/2 = s - c => BD = x = s - c => CD = a - (s - c) = a + c - s = s - b AE = AF = c + x = s Next: Area of the triangle in terms of ex-radius: Important Ex Radius Result Number 2: Ra = [ABC]/(s-a) Rb = [ABC]/(s-b) Rc = [ABC]/(s-c) Proof: [ABC] = Area of the triangle ABC. [ABC] = [CABIa] - [CBIa] [CABIa] = [CAIa] + [ABIa] [CAIa] = (base*height)/2 = (AC * DIa)/2 = (AC * Ra)/2 [ABIa] = (base*height)/2 = (AB * FIa)/2 = (AB * Ra)/2 => [CABIa] = Ra * (AB + AC)/2 [CBIa] = (BC * Ra)/2 => [ABC] = Ra * (AB + AC - B...