Given :
ABCD is a parallelogram with AP ⊥ BD and CQ ⊥ BD.
From figure,
AB || DC and BD is transversal.
(i) In Δ APB and Δ CQD,
⇒ ∠APB = ∠CQD (Both equal to 90°)
⇒ AB = CD (Opposite sides of parallelogram are equal)
⇒ ∠ABP = ∠CDQ (Alternate interior angles are equal)
∴ Δ APB ≅ Δ CQD (By A.A.S. congruence rule)
Hence, proved that Δ APB ≅ Δ CQD.
(ii) Since,
Δ APB ≅ Δ CQD
We know that,
Corresponding parts of congruent triangles are equal.
⇒ AP = CQ (By C.P.C.T.)
Hence, proved that AP = CQ.
