Balti Tee 2025 · Ülesanne 13

Let $\Omega_1=(ABP)$ and $\Omega_2=(CDP)$. Let $BC$ meet $\Omega_1$ again at $U\ne B$ and $\Omega_2$ again at $V\ne C$.

We have

$$\angle CUP=\angle BAP=\angle PCB=\angle PCU,$$

so $\triangle UPC$ is isosceles and $PU=PC$. Similarly, $PB=PV$.

Let $F$ be the midpoint of $BV$. It is also the midpoint of $CU$ and the foot of the perpendicular from $P$ to $BC$. By symmetry on the line $BC$, $F$ has equal powers with respect to $\Omega_1$ and $\Omega_2$, so $F$ lies on their radical axis.

Also,

$$EB\cdot EA=EC\cdot ED,$$

so $E$ lies on the same radical axis. The point $P$ belongs to both circles and therefore lies on the radical axis as well. Hence $P,F,E$ are collinear. Since $PF\perp BC$, it follows that $PE\perp BC$.