Estonian Mathematical Olympiad

Let $\alpha =\angle CAI=\angle IAB$, $\beta =\angle ABI=\angle IBC$, $\gamma =\angle BCI=\angle ICA$. Then $\alpha +\beta +\gamma =90^{\circ }$ and $\angle CBD=\angle CAD=\alpha =\angle DAB=\angle DCB$, yielding $$\begin{gathered} \angle KBD=\angle IBD=\alpha +\beta =90^{\circ }-\gamma , \\ \angle DCL=\angle DCI=\alpha +\gamma =90^{\circ }-\beta . \end{gathered}$$ Depending on the location of the point $K$ (Figures 50 and 51), we have either $\angle DKB=\angle DKI=\angle DCI$ or $\angle DKB=180^{\circ }-\angle IKD=\angle DCI$; in each case $\angle DKB=90^{\circ }-\beta$. Analogously, we obtain $\angle CLD=90^{\circ }-\gamma$. Hence $$\begin{gathered} \angle BDK=180^{\circ }-(90^{\circ }-\gamma )-(90^{\circ }-\beta )=\beta +\gamma =90^{\circ }-\alpha , \\ \angle LDC=180^{\circ }-(90^{\circ }-\beta )-(90^{\circ }-\gamma )=\beta +\gamma =90^{\circ }-\alpha . \end{gathered}$$ On the other hand, we have $\angle BDC=180^{\circ }-2\alpha =2(90^{\circ }-\alpha )$, meaning that both $DK$ and $DL$ bisect the angle $BDC$. Consequently, points $D,K$ and $L$ lie on a line. As the bisector of the vertex angle of the isosceles triangle $BCD$ is also the perpendicular bisector of the line segment $BC$, symmetry yields $\angle DCK=\angle KBD=90^{\circ }-\gamma$ and $\angle LBD=\angle DCL=90^{\circ }-\beta$.

a. If the point $K$ lies between points $C$ and $I$ then $$\angle DKI=\angle DKB=90^{\circ }-\beta =\angle LBD=\angle LID.$$ Hence the line $DI$ is tangent to the circumcircle of the triangle $IKL$ (as points $D,K,L$ lie on a line). If the point $K$ lies between points $I$ and $D$ then $$\angle ILD=\angle CLD=90^{\circ }-\gamma =\angle DCK=\angle DIK.$$ Analogously to the previous case, the line $DI$ must be tangent to the circumcircle of the triangle $IKL$.

b. Since $\angle DKB=90^{\circ }-\beta =\angle LBD$ and points $D,K,L$ lie on a line, the line $DB$ is tangent to the circumcircle of the triangle $BKL$. On the other hand, we have $\angle DAB=\alpha =\angle CBD=\angle EBD$, implying that $DB$ is also tangent to the circumcircle of the triangle $BEA$. Using the power of the point $D$ w.r.t. to these circles, we get $D{B}^2=DK\cdot DL$ and $D{B}^2=DA\cdot DE$, respectively. Altogether, we obtain $DA\cdot DE=DK\cdot DL$. Hence points $A,K,L,E$ are concyclic.