#### Question

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

#### Solution

Join OC.

BCD is the tangent and OC is the radius.

∴ OC ⊥ BD

⇒ `∠`OCD = 90°

⇒ `∠`OCA + `∠`ACD = 90°

But in ΔOCA

OA = OC (radii of same circle)

∴ `∠` OAC + `∠` OAC

Substituting (i)

`∠` OAC + `∠`ACD = 90°

⇒ `∠`BAC + `∠`ACD = 90°

Is there an error in this question or solution?

Solution The Given Figure Shows a Circle with Centre O and Bcd is Tangent to It at C. Show That: ∠Acd + ∠Bac = 90° Concept: Tangent to a Circle.