In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ see Fig. Show that:

(i) Δ APD ≅ Δ CQB

(ii) AP = CQ

(iii) Δ AQB ≅ Δ CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

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.