Question

Figure, in a cyclic quadrilateral ABCD diagonal AC bisects the angle C. Then prove that diagonal BD is parallel to tangent PQ of a circle which passes through the points A.

Answer 1

Given :

∠ACD = ∠ACB

Now, we can see here

that ∠PAD = ∠ABD (∵ angle in the alternate segment)

similarly ∠QAB = ∠ADB

Also AB is a common arc. ADB and ACB are the angle in the same segment.

∴ ∠ADB = ∠ACD

similarly, ∠ABD = ∠ACD

By equation (i), we find that

∠PAD = ∠ADB (alternate interior angle)

∴ PQ || BD