Let $C'$ be the reflection of $C$ in the line $AB$, and let $\omega_1$ and $\omega_2$ be the circles with centers $A$ and $B$, passing through $L$ and $K$ respectively. Since $AC' = AC = AL$ and $BC' = BC = BK$, both $\omega_1$ and $\omega_2$ pass through $C$ and $C'$. By $\angle BCA = 90^{\circ}$, $AC$ is tangent to $\omega_2$ at $C$, and $BC$ is tangent to $\omega_1$ at $C$. Let $K_1 \neq K$ be the second intersection of $AX$ and $\omega_2$, and let $L_1 \neq L$ be the second intersection of $BX$ and $\omega_1$.

By the powers of $X$ with respect to $\omega_2$ and $\omega_1$,
$$XK \cdot XK_1 = XC \cdot XC' = XL \cdot XL_1,$$
so the points $K_1, L, K, L_1$ lie on a circle $\omega_3$.
The power of $A$ with respect to $\omega_2$ gives
$$AL^2 = AC^2 = AK \cdot AK_1,$$
indicating that $AL$ is tangent to $\omega_3$ at $L$. Analogously, $BK$ is tangent to $\omega_3$ at $K$. Hence $MK$ and $ML$ are the two tangents from $M$ to $\omega_3$ and therefore $MK = ML$.