OP, OQ and QR are three straight line segments. The three parallelograms OPAQ, OQBR and ORCP are drawn. Let us prove that, AR, BP and CQ bisect each other.

In figure, OPAQ, OQBR and ORCP are three parallelograms, CQ, AR and BP are joined such that they intersect each other O.

To prove: AR, CQ and BP bisect each other i.e.

(i) OA = OR

(ii) OB = OP

(iii) OC = OQ

Now, As OQBR and ORCP are parallelograms, we have

QB || OR and QB = OR

OR || CP and OR = CP

⇒ QB || CP and QB = CP

In ΔOQB and ΔOPC

QB = CP [Proved above]

∠OQB = ∠OCP [Alternate interior angles]

∠OBQ = ∠OPC [Alternate interior angles]

ΔOQB ≅ ΔOPC [By ASA congruency criterion]

As, Corresponding parts of congruent triangles are equal, we have

OB = OP and OQ = OC

Now, As OQBR and OPAQ are parallelograms, we have

OQ || BR and OQ = BR

OQ || AP and OQ = AP

⇒ BR || AP and BR = AP

In ΔORB and ΔOPA

BR = AP [Proved above]

∠ORB = ∠OAP [Alternate interior angles]

∠OBR = ∠OPA [Alternate interior angles]

ΔORB ≅ ΔOPA [By ASA congruency criterion]

As, Corresponding parts of congruent triangles are equal, we have

OR = OA

Hence Proved!