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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.

Circles

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Answer

The figure of the parallelogram ABCD with a circle through its vertices A and B meeting side BC at point P and side AD at point Q is shown below:

In the figure, if ∠BAD = 96°, find ∠BCD and ∠BFE. Prove that AD is parallel to FE. Circles, Concise Mathematics Solutions ICSE Class 10.

We know that,

An exterior angle of a cyclic quadrilateral is equal to its opposite interior angle.

∴ ∠1 = ∠A ……. (i)

Also,

∠A = ∠C ………..(ii) [Opposite angles of a parallelogram are equal.]

From (i) and (ii) we get,

∠1 = ∠C ………(iii)

Also,

⇒ ∠C + ∠D = 180° [Sum of co-interior angles of a parallelogram = 180°]

⇒ ∠1 + ∠D = 180°

Hence, proved that PCDQ is cyclic.

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