Q23 of 63 Page 156

ABCD is a parallelogram. A circle is drawn through A and B such that it intersects AD at P and BC at Q. Prove that P, Q, C and D are cyclic.


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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