2022 CGMO

Proof. As shown in the following picture, let the extensions of $AK$ and $AL$ intersect the circle passing through $B$, $K$, $L$, $C$ at points $X$ and $Y$, respectively. Connect $BX$ and $BY$. Considering the given conditions, we have $\angle AKD=\angle BCK=\angle BXK$ and $\angle ALD=\angle BCL=\angle BYL$.

Thus, $DK\parallel BX$ and $DL\parallel BY$. So we have $$\frac{AK}{AX}=\frac{AD}{AB}=\frac{AL}{AY}.$$ Since $K$, $L$, $Y$, $X$ are concyclic, by the Power of a Point theorem, we have $$AK\cdot AX=AL\cdot AY.$$ Multiplying the two equations above, we get $A{K}^2=A{L}^2$, which implies $AK=AL$. $\square$