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Question -

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



Answer -

Given : In the figure, O is the centre of the circle BCD is a tangent, CP is a chord
 
prove : ∠BAC + ∠ACD = 90°
Proof: ∠ACD = ∠CPA (Angles in the alternate segment)
But in ∆ACP,
∠ACP = 90° (Angle in a semicircle)
∠PAC + ∠CPA = 90°
=> ∠BAC + ∠ACD = 90°
(∠ACD = ∠CPA proved)
Hence proved.

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