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Geometry Difficulty 7.7 National Olympiad, round 2 Prove it Baltic Way

Point $P$ lies inside triangle $ABC$ , $M$ is the midpoint of side $BC$ and $P'$ is symmetric to $P$ with respect to $M$ . $K$ and $H$ are projections of $P$ on $AB$ and $AC$ respectively and $KM = HM$ . Prove that $\angle PAB = \angle CAP'$ .

Solution

Denote by $A'$ , $K'$ , $H'$ the points symmetrical to $A$ , $K$ , $H$ with respect to $M$ , as in figure 13.
It is clear that $KH'H'K$ is a rectangle and the quadrilateral $AKPH$ is inscribed, denote its circumcircle by $\omega$ . Then
$\angle CHK' = 90^\circ - \angle AHK = \angle PHK = \angle PAK.$
Let $L$ be the second intersection point of line $HK'$ and $\omega$ . The above equality of angles means that $\angle LHA = \angle PAK$ , therefore $AL = KP$ , and then $AKPL$ is a rectangle. Hence $AL \parallel KP \parallel K'P'$ and $AP'K'L$ is parallelogram. Thus, $HK' \parallel AP'$ and then $\angle CHK' = \angle CAP'$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.