In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:

(i) In ΔAPD and ΔCQB,

∠ADP = ∠CBQ (Alternate interior angles for BC || AD)

AD = CB (Opposite sides of parallelogram ABCD)

DP = BQ (Given)

ΔAPD ≅ ΔCQB (Using SAS congruence rule)

(ii) As we had observed that,

 ΔAPD ≅ ΔCQB

AP = CQ (CPCT)

(iii) In ΔAQB and ΔCPD,

∠ABQ = ∠CDP (Alternate interior angles for AB || CD)

AB = CD (Opposite sides of parallelogram ABCD)

BQ = DP (Given)

ΔAQB ≅ ΔCPD (Using SAS congruence rule)

(iv) As we had observed that,

 ΔAQB ≅ ΔCPD,

AQ = CP (CPCT)

(v) From the result obtained in (ii) and (iv),

AQ = CP and AP = CQ

Since,

Opposite sides in quadrilateral APCQ are equal to each other,

APCQ is a parallelogram