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 1

Solution : In $△ODCand△OEB$ , we have:
DC = BE
$\therefore DC=AB$
$\angle COD=\angle BOE$ (Vertically opposite angles)
$\angle OCD=\angle OBE$ (Alternate interior angles)
$△ODC\cong △OEB$
⇒ OC = OB
(Cpctc )
We know that BC = OC + OB.
$\therefore EDbisectsBC$