Given : ABCD is a parallelogram and P and Q are points on diagonal BD such that DP = BQ.

**To Prove : **

(i) ∆APD ≅ ∆CQB

(ii) AP = CQ

(iii) ∆AQB ≅ ∆CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

**Proof :**

(i) In ∆APD and ∆CQB, we have

AD = BC [Opposite sides of a ||gm]

DP = BQ [Given]

∠ADP = ∠CBQ [Alternate angles]

∴ ∆APD ≅ ∆CQB [SAS congruence]

(ii) ∴ AP = CQ [CPCT]

(iii) In ∆AQB and ∆CPD, we have

AB = CD [Opposite sides of a ||gm]

DP = BQ [Given]

∠ABQ = ∠CDP [Alternate angles]

∴ ∆AQB ≅ ∆CPD [SAS congruence]

(iv) ∴ AQ = CP [CPCT]

(v) Since in APCQ, opposite sides are equal, therefore it is a parallelogram. **Proved.**