Turkish Mathematical Olympiad

Let $D'$ be the point symmetric to $D$ with respect to the line $BE$. $2 \cdot CK \cdot AC = CE \cdot AB$ implies $CK/CE = R_1/R_2$ where $R_1$ and $R_2$ are radii of the circles $\omega_1$ and $\omega_2$, respectively. By the sine law, $2 \cdot \sin \angle CLF = CK/R_1 = CE/R_2 = 2 \cdot \sin \angle CFE$, and hence $\angle CLF = \angle CFE$ since the sum of angles is less than $180^\circ$.

Let $X$ be the second intersection point of the line $BE$ with $\omega_1$. Since $[AB]$ is a diameter of $\omega_1$ and $AC = AD$, the smaller arcs $BC$ and $BD$ of $\omega_1$ are equal and hence $\angle DXB = \angle CXB$. Since $\angle D'XB = \angle DXB$, we get that the points $X, C, D'$ are collinear. Since $\angle AXB = 90^\circ$, we get $EX = FX$ and since $\angle ACB = 90^\circ$, the line $BC$ is tangent to $\omega_2$ and hence $\angle FCB = \angle BEC$. On the other hand, $\angle DCF = \angle DEF$, therefore $\angle DEC = \angle DCB = \angle CXB = \angle DXB$ and hence $\angle ECX = \angle DEX$ and

$\angle CEX = \angle EDX$.

Therefore the triangles $ECX$ and $DEX$ are similar and it follows that $XC \cdot XD' = XC \cdot XD = EX^2 = FX^2$. Since $\angle CLF = \angle CFE$, the line $XF$ is tangent to the circumcircle of triangle $CLF$ and since $XC \cdot XD' = FX^2$, the point $D'$ lies on this circle. Done.