Let $I$ be the incenter of triangle $ABC$ with $AB < AC$. Point $X$ is chosen on the external bisector of $\angle ABC$ such that $IC = IX$. Let the tangent to the circumscribed circle of $\triangle BXC$ at point $X$ intersect the line $AB$ at point $Y$. Prove that $AC = AY$.
Proposed by Oleksiy Masalitin
Lengths!
Let $Z$ be the other point on $BX$ with $IZ = IC$.
In a triangle $ABC$ with incenter $I$, we have \[IC^2 - IB^2 = BC(AC-AB).\]Proof: Let $DEF$ be the usual contact triangle, then
\[IC^2 - IB^2 = DC^2 - DB^2 = (DC+DB)(DC - DB) = BC(EC - FB) = BC(AC - AB).\]
(a) When $X$ and $C$ lie on same side of $BI$. Note that $\angle YBX = \angle ZBA = \angle XBC$ meaning $\triangle BYX \sim \triangle BXC \implies BY = \tfrac{BX^2}{BC}$.
\begin{align*}
AB + BY = AC&\iff AB + \tfrac{BX^2}{BC} = AC \\
&\iff AB\cdot BC + BX^2 = BC\cdot AC \\
&\iff IX^2 - IB^2 = BX^2 = BC(AC-AB) \\
&\iff IC^2 - IB^2 = BC(AC-AB). ~ \blacksquare
\end{align*}(b) When $X$ lies on the other side a similar solution follows with $\triangle BYZ \sim \triangle BZC$ in (a).