On the plane circles $k$ and $\ell$ are intersected at points $C$ and $D$, where circle $k$ passes through the center $L$ of circle $\ell$. The straight line passing through point $D$ intersects circles $k$ and $\ell$ for the second time at points $A$ and $B$ respectively in such a way that $D$ is the interior point of segment $AB$. Show that $AB = AC$.
Since $CL=LD$, we have $\angle CAL = \angle LAC$ (since they intercept equal arcs). Then, since also $CL=LB$, we have $\triangle CLA = \triangle CLB$ by SSA, so $AB=AC$.