Question

Please answer this and also attach the diagram made. No. 19(b)

Answer 1

Answer:

Given ABCD is a parallelogram and a circle is drawn through AB such that it intersects AD at P and BC at Q.

Here the circle passes through the points A, B, Q and P.

Hence ABQP is a cyclic quadrilateral. Therefore, ∠A + ∠BQP = 180° [Since opposite angles in a cyclic quadrilateral is supplementary] --- (1)

Also, ∠CQP + ∠BQP = 180° [linear pair] --- (2)

From (1) and (2) we get

∠A = ∠CQP --- (3)

AB || CD and AD is the transversal

Hence ∠A + ∠D = 180° [Since angles on the same side of transversal are supplementary]

⇒∠CQP + ∠D = 180° [From (3)]

Thus, in quadrilateral PQCD, opposite angles are supplementary.

Hence, quadrilateral PQCD is a cyclic quadrilateral Therefore, points, P, Q, C and D are concyclic.