Given ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q To prove Points P, Q, C and D are con-cyclic. Construction Join PQ Proof ∠1 = ∠A [exterior angle property of cyclic quadrilateral] But ∠A = ∠C [opposite angles of a parallelogram] ∴ ∠1 = ∠C ,..(i) But ∠C+ ∠D = 180° [sum of cointerior angles on same side is 180°] ⇒ ∠1+ ∠D = 180° [from Eq. (i)] Thus, the quadrilateral QCDP is cyclic. So, the points P, Q, C and D are con-cyclic. Hence proved.