Let $ABC$ be a triangle with $AB > AC$. Point $P \in (AB)$ is such that $\angle ACP = \angle ABC$. Let $D$ be the reflection of $P$ into the line $AC$ and let $E$ be the point in which the circumcircle of $BCD$ meets again the line $AC$. Prove that $AE = AC$.
Let AB meet circle at S.
∠ACD = ∠ACP = ∠ABC = ∠AES so SE = DC = PC.
∠CPA = ∠ACB = ∠ASE and ∠ACP = ∠ABC = ∠AES and SE = PC so triangles AES and APC are congruent so AE = AC.
we're Done.
$\angle ACP=\angle ABC$ So $AC^2=AP.AB$ We will show $AE^2=AP.AB$ This means we must show that $\angle ABE=\angle PEA$
If $\angle ACP=\alpha$ $\angle APD=\beta$ $\angle ADE=\theta$, we can say that $\angle PDC=90-\alpha$ so $\angle EDC=90+\beta+\theta-\alpha$. $EBCD$ is cyclic so $\angle EBC=90+\alpha+\beta-\theta$ We know that $\angle ABC=\alpha$ $\implies$ $\angle EBA=90-\beta-\theta$
$\beta=\angle ADP=\angle DPA$ and $\theta=\angle ADE=\angle APE$ $\implies$ $\angle PEA=90-\beta-\theta$