BMO 2017

Let the circumcircle of triangle $CED$ intersect $BC$ at point $G$. From power of point we have $$BG\cdot BC=BE\cdot BD$$ Combining (1) with the problem statement we get $$\frac{BG\cdot BC}{BD}=BE=\frac{B{C}^2-CD\cdot CA}{BD}$$ and from here we get $$CD\cdot CA=BC(BC-BG)=BC\cdot CG$$ (2) implies that $D$, $A$, $B$, $G$ are concyclic as well. This gives us $$\angle BEC=\angle BGD=180^{\circ }-\angle BAD=180^{\circ }-\angle CAB$$ Now let the circumcircle of $ADE$ and $BEC$ intersect again at $X$. Since $$\angle XCB=\angle XEB=180^{\circ }-\angle XED=\angle XAD=\angle XAC$$ and $$\angle BXC=\angle BEC=180^{\circ }-\angle BAC$$ we have that $X$ is on the unique circle through $A$ and $C$ tangent to side $BC$ at point $C$ and circumcircle of $BHC$ where $H$ is the orthocenter of triangle $ABC$. This intersection is unique and we are done.