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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.
Solution
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.
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