Given that,

ABCD is cyclic quadrilateral in which AB = DC

To prove: AC = BD

Proof: In and ,

AB = DC (Given)

∠BAP = ∠CDP (Angles in the same segment)

∠PBA = ∠PCD (Angles in the same segment)

Then,

(i) (By c.p.c.t)

(ii) (By c.p.c.t)

Adding (i) and (ii), we get

PA + PC = PD + PB

AC = BD