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.

ABCD is the given cyclic quadrilateral.

Also, PA = PD [Given]

So, ∠PAD = ∠PDA [As angles opposite to equal sides are equal.] ………. (1)

From figure,

∠BAD = 180° - ∠PAD [Linear pair of angles]

Similarly,

∠CDA = 180° - ∠PDA = 180° - ∠PAD [From (1)]

As sum of opposite angles of a cyclic quadrilateral = 180°,

∴ ∠ABC + ∠CDA = 180°

⇒ ∠ABC = 180° - ∠CDA

⇒ ∠ABC = 180° - (180° - ∠PAD) = ∠PAD

⇒ ∠ABC = ∠PAD.

Also,

⇒ ∠DCB + ∠BAD = 180°

⇒ ∠DCB = 180° - ∠BAD

⇒ ∠DCB = 180° - (180° - ∠PAD)

⇒ ∠DCB = ∠PAD

⇒ ∠DCB = ∠PDA [As ∠PAD = ∠PDA]

Since pairs, ∠DCB and ∠PDA, ∠ABC and ∠PAD are corresponding angles and are equal

Hence, proved that AD || BC.