Mongolian National Mathematical Olympiad

Question

Let be circumcircle of triangle ABC and let AD and BE be altitudes. A line DE intersects the circle at points P and Q with order P, E, D, Q in the line. Let bisectors of angle APQ and BQP intersect a circle at point K and L, respectively. Prove that the line KL is perpendicular to the bisector of angle ACB. (Proposed by B. Ulziinasan) FIGURE_0

Accepted Answer

Let O be a circumcenter of triangle ABC. We know BCO = 90^ - A and CDE = A so BCO + CDE = (90^ - A) + A = 90^, from here PQ CO. Hence CQ = CP and denote it by x. If we denote QB = 2y, BA = 2z, AP = 2t then BM = MA = z. Because of QK = KA we have QA = 2(y+z) and QK = y+z. From here KM = y. Because BL = LP, we have BP = 2(t+z) and LP = t+z. So we get LC = z+t+x. Now we can write KM + LC = y + z + t + x = CQ + QB + BK + KM = 180^, in other words KL MC. FIGURE_0