Question

(b) In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.​

Answer 1

Answer:

In triangles APO and OQC

∠OAP=∠OCQ ( taking Ac as transversal then angles are on opposites sides of transversal)

AP=CQ (given )

∠OPA=∠OQC ( taking PQ as transversal then angles are on opposites sides of transversal )

By ASA criterion of congruency triangle APO is congruent to triangle OQC .

PO=OQ ( byCPCT )

Answer 2

Given :

• ABCD is a parallelogram
• AP = CQ

To Prove :

• AC and PQ bisect each other

Solution :

In ∆AOP and ∆COQ

AP = CQ (given)

∠OPA = ∠OQC (Alternate interior angles are equal)

∠OAP = ∠OCQ (Alternate interior angles are equal)

∴ ∆AOP ≅ ∆COQ (By ASA Congruency)

• AO = OC (C.P.C.T)
• PO = OQ (C.P.C.T)

Hence, all length are equal

∴ AC and PQ bisect each other