Question

In the given figure, $\square$ABCD is a parallelogram. Point E is on the ray AB such that BE = AB then prove that line ED bisects seg BC at point F.

Answer 1

$\square$ABCD is a parallelogram
So, AD || CB
$\Rightarrow$AD || BF
Given: AB = BE so, B is the midpoint of AE.
By converse of mid-point theorem,
F is the midpoint of DE. So, DF = FE
In $△$EBF and $△$DCF,
DF = FE (Proved above)
DC = BE (Since DC = AB and AB = BE)
$\angle$FDC = $\angle$FEB (Alternate interior angles of the parallel lines AE and CD)
Thus, $△$EBF $\cong$ $△$DCF (SAS congruency)
Therefore, FB = FC (CPCT)
Hence, ED bisects seg BC at F.