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

Question

In fig. O is the centre of the circle and BCD is tangent to it at C. Prove that \(\angle\)BAC + \(\angle\)ACD = \(90^\circ.\)

Answer 1

Given: In the above figure, O is the centre of the circle and BCD is tangent to it at C.

To prove: ∠BAC + ∠ACD = 90°

Proof: In ΔOAC

OA = OC [radii of same circle]

⇒ ∠OCA = ∠OAC [angles opposite to equal sides are equal]

⇒ ∠OCA = ∠BAC [1]

Also,

OC ⊥ BD [Tangent at any point on a circle is perpendicular to the radius through point of contact]

⇒ ∠OCD = 90°

⇒ ∠OCA + ∠ACD = 90°

⇒ ∠BAC + ∠ACD = 90° [From 1]

Hence Proved