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,
$$\begin{align*} \angle KHL &= \angle CHB = 180^\circ - \angle BCH - \angle CBH \\ &= 180^\circ - (90^\circ - \gamma) - (90^\circ - \beta) \\ &= \beta + \gamma = 180^\circ - \alpha. \end{align*}$$
Now note (Fig. 42) that for the incenter $I$ of triangle $AKL$,
$$\begin{align*} \angle KIL &= 180^\circ - \angle IKL - \angle ILK = 180^\circ - \frac{\angle AKL}{2} - \frac{\angle ALK}{2} \\ &= 180^\circ - \frac{\angle AKL + \angle ALK}{2} \\ &= 180^\circ - \frac{180^\circ - \angle KAL}{2} = 180^\circ - \frac{180^\circ - (180^\circ - 2\alpha)}{2} = 180^\circ - \alpha, \end{align*}$$
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.

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