Question
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 $OE = OF.$
✓
Answer
Given: A parallelogram $ABCD$, in which diagonals intersect at $O$. $E$ and $F$ are the points on $AB$ and $CD$
To Prove: $\text{OE = OF}$
Proof: In $\triangle\text{AOE}$ and $\triangle\text{COF},$ we have
$\angle\text{CAE}=\angle\text{DCA}$ [Alternate angles]
$\text{AO = CO}$ [diagonals are equal and bisect each other]
and, $\angle\text{AOE}=\angle\text{COF}$ [Vertically opposite angles]
Thus by Angle-Side-Angle criterion of congruence, we have,
$\therefore\triangle\text{AOE}\cong\triangle\text{COF}$ [By ASA]
The corresponding parts of the congruent triangles are equal.
$\therefore\text{OE = OF}$ $[$By $C.P.C.T.]$
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