Question

In the figure seg ac is a diameter of the circle with centre O.bisector of angle abc intersects the circle at d.prove that seg od perpendicular to seg ac

Answer 1

Answer:  

Given : AC is the diameter of the circle with the centre C.

BD is the angle bisector of angle ABC,

That is, ∠ ABD = ∠ CBD ( by the property of angle bisector)

To prove : seg OD ⊥ seg AC

Proof :  Since, ∠ ABC = 90°  (Angle inscribed in semicircle)

⇒ ∠ ABD + ∠ CBD = 90°

⇒ ∠ ABD + ∠ ABD = 90°

⇒ 2 × ∠ ABD = 90°

⇒ ∠ ABD = 45°

Also, ∠AOD = 2 × ∠ ABD  (Central angle theorem)

⇒  ∠AOD = 2 × 45°

⇒ ∠AOD = 90°

⇒ OD ⊥ AC

Hence, proved.