Question

please solve this question ​

Answer 1

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Given ABCD is a parallelogram.

A circle through points A and B is drawn such that it intersects the AD at P and BC at Q.

We have to prove that points P, Q, C and D are cyclic.

Proof:

Since the circle passes through points A, B, P and Q, ABPQ is a cyclic quadrilateral.

We know that opposite angles in a cyclic quadrilateral are supplementary.

⇒∠A + ∠PQB = 180°

⇒∠CQP + ∠PQB = 180°

∴ ∠A = ∠CQP

Now, AB and CD are parallel lines and AD is the traversal.

We know that angles on the same side of traversal are supplementary.

⇒∠A + ∠D = 180°

⇒∠CQP + ∠D = 180°

Thus, in PQCD quadrilateral, opposite angles are supplementary.

Hence, quadrilateral PQCD is a cyclic quadrilateral.

∴ P, Q, C and D are concyclic.

Hence proved.

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