Estonian Mathematical Olympiad

Denote $\angle CAB=\alpha$, $\angle ABC=\beta$, and $\angle BCA=\gamma$. By tangency, $\angle KAH=\angle ABH=90^{\circ }-\alpha$ (Fig. 41). Since $\angle ACH=90^{\circ }-\alpha$ as well, triangles $AHC$ and $KHA$ are similar. Consequently, the corresponding third angles are equal, i.e., $\angle AKH=\angle CAH=90^{\circ }-\gamma$. Similarly, we get $\angle LAH=90^{\circ }-\alpha$ and $\angle ALH=90^{\circ }-\beta$. Thus, $\angle KAL=\angle KAH+\angle LAH=180^{\circ }-2\alpha$, with $AH$ bisecting the angle $KAL$.

On the other hand, $$\begin{array}{rl}\angle KHL& =\angle CHB=180^{\circ }-\angle BCH-\angle CBH\\ & =180^{\circ }-(90^{\circ }-\gamma )-(90^{\circ }-\beta )\\ & =\beta +\gamma =180^{\circ }-\alpha .\end{array}$$ Now note (Fig. 42) that for the incenter $I$ of triangle $AKL$, $$\begin{array}{rl}\angle KIL& =180^{\circ }-\angle IKL-\angle ILK=180^{\circ }-\frac{\angle AKL}{2}-\frac{\angle ALK}{2}\\ & =180^{\circ }-\frac{\angle AKL+\angle ALK}{2}\\ & =180^{\circ }-\frac{180^{\circ }-\angle KAL}{2}=180^{\circ }-\frac{180^{\circ }-(180^{\circ }-2\alpha )}{2}=180^{\circ }-\alpha ,\end{array}$$ so $\angle KIL=\angle KHL$. Since points $H$ and $I$ lie on the same side of line $KL$, points $K$, $H$, $I$, and $L$ lie on the same circle. Given that both $H$ and $I$ lie on the angle bisector $AH$ of angle $KAL$, which intersects the chord $KL$, it follows that $H=I$. Consequently, $\angle LKH=\angle AKH=90^{\circ }-\gamma$ and $\angle KLH=\angle ALH=90^{\circ }-\beta$. In addition, we have $\angle HLK=90^{\circ }-\beta =\angle HCB$, from which it follows that points $B$, $C$, $K$, $L$ lie on the same circle.

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Alternative solution.

As in Solution 1, we note that triangles $AHC$ and $KHA$ are similar, hence $\frac{HA}{HK}=\frac{HC}{HA}$, or $H{A}^2=HK\cdot HC$. Similarly, triangles $AHB$ and $LHA$ are similar, giving $H{A}^2=HL\cdot HB$. Therefore, $HB\cdot HL=HC\cdot HK$, from which it follows that points $B$, $C$, $K$, $L$ lie on the same circle.