IV OBM

Question

The angles of the triangle ABC satisfy A C = B A = 2. The incenter is O. K, L are the excenters of the excircles opposite B and A respectively. Show that triangles ABC and OKL are similar. FIGURE_0

Accepted Answer

AOC = 180^ - A/2 - C/2. But KAO = KCO = 90^, so AKC = A/2 + C/2. So considering triangle AKL, ALK = 90^ - A/2 - C/2 = B/2. Similarly, BLC = B/2 + C/2, so BKL = A/2. FIGURE_0 Now we use the given facts that C = A/2 and A = B/2. So OKL = BKL = C, and OLK = ALK = A. Hence triangles OKL and BCA are similar.