In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that $\angle BAC + \angle ACD = 90°$.

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Step 1: Circle with centre O, tangent BCD at C, and point A on circle
Step 2: In $	riangle OAC$: $OA = OC$ (radii of the circle)
Step 3: $\Rightarrow \angle OCA = \angle OAC$ (angles opposite equal sides)
Step 4: Now, $\angle OCD = 90°$ (radius $\perp$ tangent), so $\angle OCA + \angle ACD = 90°$

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