Question 191
Let $A'$ be the hypotenuse of the height to the side $BC$ of the acute triangle $ABC$. A disk of diameter $AA'$ intersects side $AB$ at points $A$ and $D$, and side $AC$ at points $A$ and $E$. Prove that the center of the circumscribed circle of the triangle $ABC$ lies on the support of the height at $DE$ of the triangle $ADE$.