ABCD is a parallelogram. AC and BD are the diagonals intersect at O. P and Q are the points of tri section of the diagonal BD. Prove that CQ || AP and also AC bisects PQ (see figure).

Given:- ABCD is parallelogram

BP=BD/3

DQ=BD/3

Formula used:- SAS congruency property

If 2 sides and angle between the two sides of both

the triangle are equal, then both triangle are congruent

Solution:-

⇒ In parallelogram ABCD

AO=OC and DO=OB [diagonal of parallelogram bisect each other]

As DO=OB

Where DO=DQ+OQ

OB=OP+PB

∴ DQ+OQ=OP+PB

⇒ +OQ=OP+

⇒ OQ=OP+ -

⇒ OQ=OP

PQ = OP+OQ = 2(OQ) = 2(OP)

∴ AC diagonal bisect PQ

⇒ In Δ QOC and Δ POA

OQ=PO [Proven above]

AO=OC [Diagonal of parallelogram bisect each other]

∠ QOC=∠ AOP [vertically opposite angles]

Hence Δ QOC ≅ Δ POA

∴ ∠ CQO=∠ OPA

If QC and PA are 2 lines

And QP is the transverse

And ∠ CQO=∠ OPA by Alternate angles

∴ QC||PA

Conclusion:-

CQ||PA and CA is bisector of PQ.