Comparing angles formed by tangents from different points

Join $OA$ and $OC$. Since $AB$ and $CD$ are tangents at $A$ and $C$ respectively, $OA \perp AB$ and $OC \perp CD$, so $\angle OAB = 90^\circ$ and $\angle OCD = 90^\circ$. In $\triangle OAC$, $OA = OC$ (radii of same circle), so $\angle OAC = \angle OCA$. Now, $\angle BAC = 90^\circ - \angle OAC$ (or $90^\circ + \angle OAC$ depending on direction) and $\angle DCA = 90^\circ - \angle OCA$. Since $\angle OAC = \angle OCA$, we get $\angle BAC = \angle DCA$. Yes, $\angle BAC = \angle DCA$.