Balti Tee 2018 · Ülesanne 13

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