58th Ukrainian National Mathematical Olympiad

Question

There is isosceles obtuse triangle ABC with vertex in point B given. Perpendicular bisector to side BC intersects lines AC and AB in points K and M respectively. Prove, that point, symmetric to point A with respect to line BK, is on line CM. (Anton Trigub)

Accepted Answer

Let A_1 be point, symmetric to A with respect to BK (Fig. 37). At first, BA_1K = BAK = BCK, thus quadrilateral BA_1CK is cyclic. Then, A_1CB = A_1KB = AKB = 2 BCA = MBC = MCB, So we get, that points C, A_1, M are on one line.