Given that □ABCD is a parallelogram.
BD is a diagonal.
AP ⊥ BD and CQ ⊥ BD
i) In ΔAPB and ΔCQD
AB = CD ( ∵ Opp. sides of //gm ABCD)
∠APB = ∠CQD (each 90°)
∠PBA = ∠QDC (alt. int. angles for the lines AB and DC)
∴ ΔAPB ≅ ΔCQD (AAS congruence)
ii) From (1) ΔAPB ≅ ΔCQD
⇒ AP = CQ (CPCT)