Olympiad level | Similar Triangles

Let $E$ be a center of the circle $KCL$, $F$ be an intersection points of $AD$ with the circle $BDC$,

which different from $D$ and $\measuredangle BAK=\alpha$.

Thus, $$\measuredangle FBC=\measuredangle BCD=2\alpha$$ and since $$BK=BA=CD=BF,$$ we obtain $$\measuredangle BKF=90^{\circ}-\alpha.$$

In another hand, $$\measuredangle KEC=2\measuredangle KLC=2\alpha,$$ which gives $$\measuredangle EKC=90^{\circ}-\alpha,$$ which says points $E$, $K$ and $F$ are placed on the same line $FE$.

But $$\measuredangle FBC=\measuredangle FEC=2\alpha,$$ which says that $E$ is placed on the circle $BDC$ and we are done!