In the adjoining figure, ABCD is a parallelogram whose diagonals intersect each other at O. A line segment EOF is drawn to meet AB at E and DC at F. Prove that OE = OF.

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Solution

In ∆ODF and ∆OBE, we have:
OD = OB (Diagonals bisects each other)
∠DOF = ∠BOE (Vertically opposite angles)
∠FDO = ∠OBE (Alternate interior angles)
i.e., ∆ODF ≅ ∆OBE
∴ OF = OE (CPCT)
Hence, proved.

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In the adjoining figure, ABCD is a parallelogram whose diagonals intersect each other at O. A line segment EOF is drawn to meet AB at E and DC at F. Prove that OE = OF.

In ∆​ODF and ∆​OBE, we have: OD = OB (Diagonals bisects each other) ∠DOF = ∠BOE (Vertically opposite angles) ∠FDO = ∠OBE (Alternate interior angles) i.e., ∆​ODF ≅ ∆​OBE ∴​ OF = OE (CPCT) Hence, proved.

In ∆ODF and ∆OBE, we have:
OD = OB (Diagonals bisects each other)
∠DOF = ∠BOE (Vertically opposite angles)
∠FDO = ∠OBE (Alternate interior angles)
i.e., ∆ODF ≅ ∆OBE
∴ OF = OE (CPCT)
Hence, proved.