In fig.. O is the center of the circle and BCD is tangent to it at C. Prove that ∠BAC +

∠ACD = 90°

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#### 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

Is there an error in this question or solution?

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