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Question
ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.
Solution
Let ABCD be the given cyclic quadrilateral
Also, PA = PD ...(Given)
∴ ∠PAD = ∠PDA ...(1)
∴ ∠BAD = 180° – ∠PAD
And ∠CDA = 180° – PDA
= 180° – ∠PAD ...(From (1))
We know that the opposite angles of a cyclic quadrilateral are supplementary
∴ ∠ABC = 180° – ∠CDA
= 180° – (180° – ∠PAD)
= ∠PAD
And ∠DCB = 180° – ∠BAD
= 180° – (180° – ∠PAD)
= ∠PAD
∴ ∠ABC = ∠DCB = ∠PAD = ∠PAD
That means AD || BC
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