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Ukrainian National Mathematical Olympiad

Geometry Difficulty 4.9 AIME Prove it Ukraine

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

(Anton Trygub)

Solution

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

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