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

CBSE Class 10 Maths PYQ · Circles · Triangle & Circle · 3 Marks · March 2025 · Standard

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973 Marks · March 2025 · Standard

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

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In $\Delta OAC, OA = OC \implies \angle OCA = \angle OAC$. Now, $\angle OCD = 90^\circ \implies \angle OCA + \angle ACD = 90^\circ \implies \angle OAC + \angle ACD = 90^\circ$ or $\angle BAC + \angle ACD = 90^\circ$.