In the figure, $ O $ is the centre of the circle and $ B C D $ is tangent to it at $ C $. Prove that $ \angle B A C+\angle A C D=90^{\circ} $."

Given:

In the figure, \( O \) is the centre of the circle and \( B C D \) is tangent to it at \( C \).

To do:
We have to prove that \( \angle B A C+\angle A C D=90^{\circ} \).

 Solution:

From the figure,

$\angle ACD = \angle CPA$     (Angles in the alternate segment are equal)

In $\triangle ACP$,

$\angle ACP = 90^o$    (Angle in a semicircle)

$\angle PAC + \angle CPA = 90^o$

$\angle BAC + \angle ACD = 90^o$     ($ACD = ∠CPA$)

Hence proved.