Let $O$ be the center of $(CDE)$. We have $\angle BCO=\angle ECO=90^{\circ}-\angle CDE=90^{\circ}-\angle CBA$, so $CO\perp AB$. So $CO$ is the perpendicular bisector of $XY$ and also the altitude of $\triangle ABC$.

$\angle CXE=\angle CDE=\angle CBX$ hence CX^2=CE*CB and CY^2=CD*CA so CX=CY

we know that $\angle {CYD} = \angle{CED} =\angle{CAB}$ so $CY$ is touch the circle of $ADY$ and other $CX$ to $XEB$ so: $CY^2=CD.CA$ $CX^2=CE.CB$ we know that: $CD.CA=CE.CB$ so: $CY^2=CX^2$ so $CY=CX$!

But this is a well known property! $A, B, D, E$ cyclic, means that an altitude of a triangle ($\triangle ABC$) is the radius from $C$ of the other one! Best regards, sunken rock

Dear Mathlinkers, 1. by the Reim's theorem, AB is parallel to the tangent to (CDE) at C 2. O being the center of this circle, OC is perpendicular to XY in his midpoint. Sincerely Jean-Louis

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%203.pdf p. 45... Sincerely Jean-Louis

Let $O$ be the circumcenter of triangle $ADE$. It suffices to show that $O$ lies on the $A$-altitude, or $\angle DAO = 90^\circ - B$. But $$\angle DAO = 90^\circ - \frac 12 \angle AOD = 90^\circ - \angle AED = 90^\circ - B$$as $DECB$ is cyclic, so we are done.

This was in 106 geometry problems by Titu Andreescu.