I was asked to prove the following problem.
Consider the following diagram where a triangle $ABC$ lies inside its circumcircle, $D$ is the point where the angle bisector $\alpha$ of $B$ intersects $AC$ and $E$ the point where the angle bisector $\beta$ of $C$ intersects $AB$. The points $P$ and $Q$ are the intersection of the line $DE$ with the circle.
We wish to show that $AP = AQ \iff AB = AC$.
It is readily seen from the picture that the triangle $ABC$ must be isosceles in order for the relation to be true. The hard part here is to show that it is indeed the case only under the assumption that $AP = AQ$.
It was suggested to me to first show that $AP^2 = AE \cdot AB$ and similarly, $AQ^2 = AD \cdot AC$. Equating these should reveal that $ABC$ is similar to $ADE$. Following this, one can then deduce that the quadrilateral $BCDE$ is cyclic (as shown in the picture), so that $ABC$ is isosceles, i.e. $AB = AC$.
The problem is that I cannot even show that very first part, namely $AP^2 = AE \cdot AB$. It seems to me that this holds if and only if rotating anticlockwise the segment $AP$ onto $AB$ with pivot $A$ and intersecting $P$ with $AB$ at $P'$ gives that $P'$ is the midpoint of $B$ and $E$. However, I don't know how to prove this and this doesn't look like a trivial exercise.
One thing I also realized is that the triangles $ABP$ and $AEP$ should be similar. Showing this should suffice to prove the equality $AP^2 = AE \cdot AB$. Working from there, I should be able to find corresponding angles in $BCDE$ and show that it is cyclic.
There might be a quicker and easier approach to this problem, but so far I have found none. Any help or hint would be greatly appreciated.
From
(1) $\angle PAE$ is the common angle.
(2) $\angle ABP = \angle Q = \angle P$ [equal chord, equal angles] + [base angles, isosceles ⊿]
∴⊿APE~⊿ABP
Thus, $AP^2 = AB. EA$