We examine the configuration in the figure; other configurations proceed analogously. It holds that $\angle I D B=90^{\circ}=\angle I K B$, so $B K D I$ is a cyclic quadrilateral. Furthermore, $\angle A L B=90^{\circ}=\angle A K B$, so $B K L A$ is also a cyclic quadrilateral. Therefore,

$$\angle B K D=180^{\circ}-\angle B I D=180^{\circ}-\left(90^{\circ}-\frac{1}{2} \angle A B C\right)=180^{\circ}-\angle B A L=\angle B K L.$$

From this, it follows that $K, D$, and $L$ lie on a straight line. Analogously, $K, E$, and $L$ also lie on a straight line, and this is the same line, so $D$ lies on it as well.
Let $S$ be the second intersection point of the circumcircles of $\triangle D K H$ and $\triangle E L H$. Then it holds that

$$\begin{gathered} \angle D S E=360^{\circ}-\angle D S H-\angle H S E=\angle D K H+180^{\circ}-\angle H L E=\angle L K H+\angle H L K \\ =180^{\circ}-\angle K H L . \end{gathered}$$

Since $H L I K$ is also a cyclic quadrilateral (due to two right angles), it holds that
$180^{\circ}-\angle K H L=\angle K I L=\angle A I B=180^{\circ}-\angle I B A-\angle I A B=180^{\circ}-\frac{1}{2} \angle C B A-\frac{1}{2} \angle C A B$.
Thus, $\angle D S E=180^{\circ}-\frac{1}{2} \angle C B A-\frac{1}{2} \angle C A B$. Now let $F$ be the tangency point of the incircle with side $A B$, then $A F I E$ and $B F I D$ are also cyclic quadrilaterals (both due to two right angles). Therefore, it holds that

$$\angle D F E=\angle D F I+\angle I F E=\angle D B I+\angle I A E=\frac{1}{2} \angle C B A+\frac{1}{2} \angle C A B.$$

We conclude that $\angle D F E+\angle D S E=180^{\circ}$, which implies that $S$ lies on the circumcircle of $\triangle D E F$, which is precisely the incircle of $\triangle A B C$.
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