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Geometry Difficulty 5.1 AIME, harder Prove it Turkey

Let $ABC$ be an acute triangle, $P$ be the midpoint of the side $BC$ and $K$ be the foot of the altitude from $A$ . Let $D$ be a point on the segment $AP$ such that $\angle BDC = 90^\circ$ . Let the second intersection point of the circumcircle of $ADK$ and line $BC$ be $E$ . Let the second intersection point of the circumcircle of $ABC$ and line $AE$ be $F$ . Prove that $\angle AFD = 90^\circ$ .

Solution

Since $A$ , $D$ , $K$ , $E$ are concyclic we have
$\angle AKD = \angle ADE = 90^\circ$
and hence $\triangle ADE$ is a right triangle. Since $D$ lies on the circle centered at $P$ and $ED \perp PD$ , we can see that $ED$ is tangent to the circumcircle of $BDC$ . Using the power of the point $E$ with respect to the circles $(BDC)$ and $(ABC)$ we get
$ED^2 = EB \cdot EC = EF \cdot EA$
and from the Euclid relations in the triangle $ADE$ we are done.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.