Question

ABCD is a parallelogram. A circle through vertices A and B meets side BC at point P and side AD at point Q. Show that quadrilateral PCDQ is cyclic.

Answer 1

Given: ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q. To prove points P, Q, C and D are con-cyclic.

Construction: Join PQ 

Proof: ∠1 = ∠A  ...[Exterior angle property of cyclic quadrilateral]

But ∠A = ∠C  ...[Opposite angles of a parallelogram]

∴ ∠1 = ∠C   ...(i)

But ∠C + ∠D = 180°   ...[Sum of cointerior angles on same side is 180°]

`=>` ∠1 + ∠D = 180°  ...[From equation (i)]

Thus, the quadrilateral QCDP is cyclic.

So, the points P, Q, C and D are con-cyclic.

Hence proved.