Question

In parallelogram ABCD, the angle bisector of $\angle A$ bisects BC. Will angle bisector of B also bisect AD? Give reason.

Answer 1

Given, ABCD is a parallelogram, bisector of $\angle A$, bisects BC at F, i.e. $\angle 1=\angle 2$, CF = FB.
Draw $FE\parallel BA$
ABFE is a parallelogram by construction [$\because FE\parallel BA$]
$\Rightarrow \angle 1=\angle 6$ [alternate angle]

But $\angle 1=\angle 2$ [given]
$\therefore \angle 2=\angle 6$
AB = FB [opposite sides to equal angles are equal] ..... (i)
$\therefore ABFEisarhombus$.
Now, in $\Delta ABO$ and $\Delta BOF$, AB = BF [from Eq. (i)]
BO = BO [common]
AO = FO [diagonals of rhombus bisect each other]
$\therefore \Delta ABO\cong \Delta BOF$ [by SSS]
$\angle 3=\angle 4$ [by CPCT]
Now, $BF=\frac{1}{2}BC$ [given]
$\Rightarrow BF=\frac{1}{2}AD$ [BC = AD]
$\Rightarrow AE=\frac{1}{2}AD$ [BF = AE]
$\therefore$ E is the mid-point of AD.