The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that : ∠ACD + ∠BAC = 90°.

Join OC.

We know that,

The radius from the center of the circle to the point of tangent is perpendicular to the tangent line.

BCD is the tangent and OC is the radius.

As, OC ⊥ BD

∠OCD = 90°

∴ ∠OCA + ∠ACD = 90° …………. (1)

In ∆OCA,

⇒ OA = OC [Radius of the same circle]

∴ ∠OCA = ∠OAC [As, angles opposite to equal sides are equal]

Substituting in (1), we get

⇒ ∠OAC + ∠ACD = 90°

⇒ ∠BAC + ∠ACD = 90° [From figure, ∠BAC = ∠OAC]

Hence, proved that ∠ACD + ∠BAC = 90°.