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Ukrainian National Mathematical Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it Ukraine

In triangle $ABC$ , perpendicular bisector of the side $AC$ intersects the angle bisector $AK$ in point $P$ , $M$ is such point that $\angle MAC = \angle PCB$ , $\angle MPA = \angle CPK$ , and points $M$ and $K$ lie on different sides from the segment $AC$ . Prove that the line $AK$ divides the segment $BM$ in two equal segments.

Solution

Let $T$ be the point, symmetrical to $M$ with respect to $AK$ (Fig. 4). Obviously, to prove the statement, it suffices to prove that $BT \parallel AK$ . Notice that points $C, P, T$ lie on the same line. Also,
$\angle TAB = \angle TAK - \angle BAK = \angle MAK - \angle KAC = \angle CAM = \angle TCB,$
Which means the quadrilateral $BCAT$ is inscribed, and
$\angle TBA = \angle TCA = \angle PAC = \angle PAB,$
which proves the parallelism.

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