Question

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

Answer 1

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)

∴ ∠OCA + ∠OAC 

Substituting (i)

∠OAC + ∠ACD = 90°

`=>` ∠BAC + ∠ACD = 90°