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Geometry Difficulty 8.0 National Olympiad, round 2 Prove it Baltic Way

$AD$ is a bisector of the triangle $ABC$ . Line $AD$ intersects a second time the circumcircle of $\triangle ABC$ at point $E$ . Let $K, L, M$ and $N$ be the midpoints of the segments $AB, BD, CD$ and $AC$ respectively, $P$ be the circumcenter of the triangle $EKL$ , $Q$ be the circumcenter of the triangle $EMN$ . Prove that $\angle PEQ = \angle BAC$ .

Solution

Triangles $AEB$ and $BED$ are similar since $\angle BAE = \angle EAC = \angle DBE$ . Hence $\angle AEK = \angle BEL$ as the angles between a median and a side in similar triangles. Denote these angles by $\varphi$ . Then $\angle EKL = \varphi$ since $KL$ is a midline of $\triangle ABD$ .

Analogously, let $\psi = \angle AEN = \angle CEM = \angle ENM$ . And let $\beta = \angle ABC$ , $\gamma = \angle ACB$ .

The triangle $PEL$ is isosceles, therefore $\angle PEL = 90^\circ - \frac{1}{2}\angle EPL = 90^\circ - \angle EKL = 90^\circ - \varphi$ and

$\angle PEA = \angle PEL - \angle AEL = \angle PEL - (\angle AEB - \angle BEL) = 90^\circ - \varphi - (\gamma - \varphi) = 90^\circ - \gamma.$

Analogously $\angle QEA = 90^\circ - \beta$ .

Thus $\angle PEQ = \angle PEA + \angle QEA = 180^\circ - \beta - \gamma = \angle BAC$ .

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