58th Ukrainian National Mathematical Olympiad

Question

Given quadrilateral ABCD, such that AB = BC. K is the midpoint of CD, rays BK and AD intersect at M, the circumcircle of ABM intersects line AC for the second time at point P. Show that BKP = 90^. (Anton Trygub)

Accepted Answer

Let N be the midpoint of diagonal AC, then NK is the midline of ACD (Fig. 31). Thus, (KN, NP) = (AM, AP) = (BM, BP) = (KB, BP), hence BNKP is inscribed. Thus, since ABC is equilateral, BNP = 90^, therefore, BKP = 90^.