Answer :

** Given:** two intersecting circles, which intersect each other at P and Q, A quadrilateral ABCD is drawn, such that sides AC and BD are passing through P and Q. Also, AC = BD

** To prove:** ABCD is cyclic

Construction: Join PQ, and extend AC and BD to meet at R.

Proof:

As, APBQ is a cyclic quadrilateral

∠BAP + ∠BQP = 180°

⇒ ∠BQP = 180° – ∠1 …[1]

Also, By linear pair

∠BQP + ∠PQD = 180°

⇒ 180° – ∠1 + ∠PQD = 180°

⇒ ∠PQD = ∠1

Also, CDQP is also a cyclic quadrilateral

⇒ ∠PQD + ∠PCD = 180°

⇒ ∠1 + ∠PCD = 180°

⇒ ∠PCD = 180° – ∠1 …[2]

Similarly, we can prove

∠QDC = 180° – ∠2 …[3]

Also, By linear pair

⇒ ∠PCD + ∠RCD = 180°

⇒ 180 – ∠1 + ∠RCD = 180° [From 2]

⇒ ∠RCD = ∠1

Hence, AB || CD [As, corresponding angles are equal through transversal AR]

Now,

In ΔABR, AB || CD and CD intersects AR and BR, therefore by basic proportionality theorem

As, AC = BD is given, we have

AR = BR

⇒ ∠1 = ∠2 [Angles opposite to equal sides are equal] …[4]

Now,

∠A + ∠D

= ∠1 + ∠QDC [From 2]

= ∠1 + 180° – ∠2

= ∠1 + 180° – ∠1 [From 4]

= 180°

Hence, ABCD is cyclic.

[As in a cyclic quadrilateral, sum of any pair of opposite angles is 180°].

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