Given :
ABCD is a parallelogram.
From figure,
AD || BC and BD is transversal
⇒ ∠ADP = ∠CBQ (Alternate angles are equal)
(i) In Δ APD and Δ CQB,
⇒ AD = CB (Opposite sides of parallelogram ABCD are equal)
⇒ DP = BQ (Given)
⇒ ∠ADP = ∠CBQ (Proved above)
∴ Δ APD ≅ Δ CQB (By S.A.S. congruence rule)
Hence, proved that Δ APD ≅ Δ CQB.
(ii) In Δ APD ≅ Δ CQB,
We know that,
Corresponding parts of congruent triangles are equal.
⇒ AP = CQ (By C.P.C.T.) ........(1)
Hence, proved that AP = CQ.
(iii) In Δ AQB and Δ CPD,
⇒ AB = CD (Opposite sides of parallelogram ABCD are equal)
⇒ ∠ABQ = ∠CDP (Alternate interior angles are equal)
⇒ BQ = DP (Given)
⇒ Δ AQB ≅ Δ CPD (By S.A.S. congruence rule)
Hence, proved that Δ AQB ≅ Δ CPD.
(iv) Since Δ AQB ≅ Δ CPD,
We know that,
Corresponding parts of congruent triangles are equal.
⇒ AQ = CP (C.P.C.T.) ............(2)
Hence, proved that AQ = CP.
(v) From equation (1) and (2), we get :
⇒ AQ = CP and AP = CQ
Since both pairs of opposite sides in APCQ are equal,
Hence, proved that APCQ is a parallelogram.