Question

In fig. 10.55, O is the centre of the circle and BCD is tangent to it at C. Prove that .

Answer 1

The question is incomplete. The complete question is,

In fig. 10.55, O is the centre of the circle and BCD is tangent to it at C. Prove that ∠BAC  + ∠ACD = 90°

Solution:

OA = OC ( radii of same circle)

=> ∠OAC = ∠OCA .... (1)

Tangents at any point of the circle is perpendicular to the radius at the point of contact.

Therefore, ∠OCA = 90°

=> ∠ACD + ∠OCA =  90°

=> ∠ACD + ∠OAC =  90° ... ( from (1))

=>  ∠ACD +  ∠BAC = 90°

Hence proved.