Question
In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC.
✓
Answer
Given: ABCD is a parralegram in which AB is produced to E such that BE = AB. DE is joined which cuts BC at O.
To Prove: $\text{OB = OC}$
Proof: In $\triangle\text{OCD}$ and $\triangle\text{OBE},$ we have,
$\angle\text{DOC}=\angle\text{EOB}$ [vertically opposite angles are equal]
$\angle\text{OCD}=\angle\text{OBE}$ [AB || CD, BC is a transversal thus, alternate angles are equal]
$\text{DC = BE}$ [AB = CD and BE = AB]
Thus, by Angle-Angle-Side criterion of congruence, we have
$\therefore\triangle\text{OCD}\cong\triangle\text{OBE}$ [by AAS]
The corresponding parts of the congruent triangles are equal.
$\therefore\text{OC = OB}$
Hence, ED bisect BC.
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