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In the given figure, $AB$ and $CD$ are tangents to a circle centred at $O$. Is $\angle BAC = \angle DCA$? Justify your answer.
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Join $OA$ and $OC$
$OA = OC$
$\angle OAC = \angle OCA$
Also, $\angle OAB = \angle OCD$
$\Rightarrow \angle OAC + \angle OAB = \angle OCA + \angle OCD$
$\Rightarrow \angle BAC = \angle DCA$
$OA = OC$
$\angle OAC = \angle OCA$
Also, $\angle OAB = \angle OCD$
$\Rightarrow \angle OAC + \angle OAB = \angle OCA + \angle OCD$
$\Rightarrow \angle BAC = \angle DCA$