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Question
In fig.. O is the center of the circle and BCD is tangent to it at C. Prove that ∠BAC +
∠ACD = 90°
Solution
Given
O is center of circle
BCD is tangent.
Required to prove: ∠BAC + ∠ACD = 90°
Proof: OA = OC [radius]
In ΔOAC, angles opposite to equal sides are equal.
∠OAC = ∠OCA …. (i)
∠OCD = 90° [tangent is radius are perpendicular at point of contact]
∠ACD + ∠OCA = 90°
∠ACD + ∠OAC = 90° [∵ ∠OAC = ∠BAC]
∠ACD + ∠BAC = 90° ⟶ Hence proved
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