DC = AB ………. (1)
AD = CB
QC = AP ……. (2) (as ABCD is a parallelogram)
(1) – (2) ⇒ DC – QC = AB – AP;
∴ DQ = PB When, ∆ APD, ∆ CQB are considered.
AD = CB
AP = QC
∠A = ∠C, The two sides and angle formed by them in these triangles are equal. So the third sides PD and BQ are equal.
∴ Two pairs of opposite sides in the quadrilateral PBQD are equal. So PBQD is a parallelogram.