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

Solution:
In $△ O D F a n d △ O B E$ , we have:
OD = OB
(Diagonals bisects each other)
$∠ D O F = ∠ B O E$
(Vertically opposite angles)
$∠ F D O = ∠ O B E$
(Alternate interior angles)
i.e., $△ O D F ≅ △ O B E$
∴ OF = OE
(Cpctc )
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.

Solution: In △ODFand△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 (Cpctc ) Hence, proved.

Solution:
In $△ O D F a n d △ O B E$ , we have:
OD = OB
(Diagonals bisects each other)
$∠ D O F = ∠ B O E$
(Vertically opposite angles)
$∠ F D O = ∠ O B E$
(Alternate interior angles)
i.e., $△ O D F ≅ △ O B E$
∴ OF = OE
(Cpctc )
Hence, proved.