Question

ABCD is a parallelogram. AC and BD are the diagonals intersect at ‘O’. P and Q are the points of trisection of the diagonal BD. Prove that CQ//AP and also AC bisects PQ.

Answer 1

Given □ABCD is a parallelogram; 

BD is a diagonal. 

P, Q are the points of trisection of BD. 

In ΔAPB and ΔCQD 

AB = CD (•.• Opp. sides of //gm ABCD) 

BP = DQ (given) 

∠ABP = ∠CDQ (alt. int. angles for the lines 

AB//DC, BD as a transversal) 

ΔAPB = ΔCQD (SAS congruence) 

Similarly in ΔAQD and ΔCPB 

AD = BC (opp. sides of //gm ABCD) 

DQ = BP (given) 

∠ADQ = ∠CBP (all int. angles for the lines AD//BC, BD as a transversal)

ΔAQD ≅ ΔCPB 

Now in □APCQ 

AP = CQ (CPCT of AAPB, ACQD) 

AQ = CP (CPCT of AAQD and ACPB) 

∴ □APCQ is a parallelogram. 

∴ CQ//AP (opp. sides of//gm APCQ) 

Also AC bisects PQ. [ ∵ diagonals of //gm APCQ]