Estonian Math Competitions

Answer: $90^{\circ }$.

Denote $\angle CAB=\alpha$ and $\angle BCA=\gamma$. Then $\angle BDC=\gamma$ and triangles CBD and ABC will be similar due to having two equal angles (see figure below).

Therefore $\angle BCD=\alpha$ and $\angle BKD=2\alpha$. But then $\angle KBC=2\alpha -\alpha =\alpha =\angle KCB$, which yields $KB=KC$. However $KC=KD$ by the choice of $K$, which yields $\angle KBD=\angle KDB=\gamma$.

Now triangle $KBD$ gives us $\gamma +\gamma +2\alpha =180^{\circ }$ or $\alpha +\gamma =90^{\circ }$.

Finally, since two angles in triangle $ABE$ are $\alpha$ and $\gamma$, the third angle $AEB$ must be $180^{\circ }-(\alpha +\gamma )=90^{\circ }$.