Given an isosceles triangle $ABC$ with base $BC$. Inside the side $BC$ is given a point $D$. Let $E, F$ be respectively points on the sides $AB, AC$ that $|\angle BED | = |\angle DF C| > 90^o$ . Prove that the circles circumscribed around the triangles $ABF$ and $AEC$ intersect on the line $AD$ at a point different from point $A$.
(Patrik Bak, Michal Rolínek)
Let $(AEC)$ and $(ABF)$ intersect $BC$ again at $P$ and $Q$ respectively. Note that $\angle PBE=\angle CBA=\angle ACB=\angle ACP=\angle BEP \iff PB=PE$ and $\angle PED=90^{\circ} - \angle BEP=90^{\circ}-\angle PBE=90^{\circ}-\angle DBE=\angle EDB=\angle EDP \iff PE=PD$, so $PB=PD$. Similarly, we can show $QC=QD$. Then $Pow_{(AEC)}D=DP \cdot DC=\frac{1}{2}DB \cdot DC=DB \cdot DQ=Pow_{(ABF)}D$. So $D$ lies on the radical axis of the two circles. Since the radical axis is the line connecting the two intersections of the circles, their second intersection must lie on line $AD$.
Define $f(X) = Pow(X,(ABF)) - Pow(X,(ACE))$, for all point $X$ on the plane. It's well known that $f$ is linear.
Claim 1: $\triangle DBE \sim \triangle DCF$.
Proof: Obviously $\angle DBE = \angle DCF$ and $\angle BED = \angle CFD$.
Hence by Claim 1, $BD/BE = CD/CF \iff BD \cdot CF = BE \cdot CD$. Then we have
\begin{align*}
f(D) &= \frac{BD \cdot f(C) + CD \cdot f(B)}{BC}\\
&= \frac{BD \cdot CF \cdot CA - CD \cdot BE \cdot BA}{BC}\\
&= \frac{BA}{BC} \cdot (BD \cdot CF - CD \cdot BE)\\
&= 0
\end{align*}Thus $D$ lies on the radical axis of $(ABF)$ and $(AEC)$.