Question

$\square$ABCD is a parallelogram. Take point E on the side AB, such than BE = AB. Prove that ED bisects BC.

Answer 1

Given:

$\square$ ABCD is a parallelogram in which E is a point on the produced side AB such that BE = AB

To prove: ED bisects BC, i.e., OB = OC
Proof:
In parallelogram ABCD, side AB || side CD, AB = CD and seg BC is a transversal.
Thus, we have:
$\angle$DCB = $\angle$CBE [Alternate angles] …(1)
And, we have:
BE = AB
$\Rightarrow$BE = CD …(2)
Now, in ΔOBE and ΔOCD, we have:
$\angle$BOE = $\angle$DOC [Vertically opposite angles]
$\angle$OBE = $\angle$OCD [From (1)]
BE = CD [From (2)]
Thus, by AAS congruency criterion, we have:
ΔOBE $\cong$ ΔOCD
$\Rightarrow$OB = OC [By c.s.c.t]
In other words, O is the midpoint of BC.
Hence, ED bisects BC.