Problem

Source: USAJMO 2023/6

Tags: AMC, USA(J)MO, USAJMO, geometry, xtimmyGgettingflamed



Isosceles triangle $ABC$, with $AB=AC$, is inscribed in circle $\omega$. Let $D$ be an arbitrary point inside $BC$ such that $BD\neq DC$. Ray $AD$ intersects $\omega$ again at $E$ (other than $A$). Point $F$ (other than $E$) is chosen on $\omega$ such that $\angle DFE = 90^\circ$. Line $FE$ intersects rays $AB$ and $AC$ at points $X$ and $Y$, respectively. Prove that $\angle XDE = \angle EDY$. Proposed by Anton Trygub