Question
In parallelogram $ABCD$, the angle bisector of $\angle\text{A}$ bisects $BC$. Will angle bisector of $B$ also bisect $AD$? Give reason.
✓
Answer
Given, $ABCD$ is a parallelogram, bisector of $\angle\text{A}$, bisects $BC$ at F i.e., $\angle\text{1}=\angle2,\text{CF}=\text{FB}$ Draw $FB\ ||\ BA$
$\therefore\text{ABFE}$ is a parallelogram by construction
$\angle1=\angle6$
But $\angle1=\angle2$ [given]
$\therefore\angle2=\angle6$
$AB = FB$ [opposite sides to equal angles are equal] $…(i)$
$\therefore\text{ABFE}$ is a rhombus.
Now, in $\triangle\text{ABO}\ \text{and}\ \text{BOF},\text{AB}=\text{FB}$ [from eq. $(i)$]
$BO = BO$ [common] $AO = FO$ [diagonals of rhombus bisect each other]
$\therefore\triangle\text{ABO}\cong\triangle\text{BOF}$ [by SSS]
$\angle3=\angle4$ [by $CPCT$]
Now, $BF =$$\frac{1}{2}\text{BC}$[given]
$\Rightarrow\text{BF}=\frac{1}{2}\text{AD}$ $[BC = AD]$
$\Rightarrow\text{AE}=\frac{1}{2}\text{AD}$$[BF = AE]$
$\therefore$ $E$ is the mid-point of $AD$.
$\therefore\text{ABFE}$ is a parallelogram by construction
$\angle1=\angle6$
But $\angle1=\angle2$ [given]
$\therefore\angle2=\angle6$
$AB = FB$ [opposite sides to equal angles are equal] $…(i)$
$\therefore\text{ABFE}$ is a rhombus.
Now, in $\triangle\text{ABO}\ \text{and}\ \text{BOF},\text{AB}=\text{FB}$ [from eq. $(i)$]
$BO = BO$ [common] $AO = FO$ [diagonals of rhombus bisect each other]
$\therefore\triangle\text{ABO}\cong\triangle\text{BOF}$ [by SSS]
$\angle3=\angle4$ [by $CPCT$]
Now, $BF =$$\frac{1}{2}\text{BC}$[given]
$\Rightarrow\text{BF}=\frac{1}{2}\text{AD}$ $[BC = AD]$
$\Rightarrow\text{AE}=\frac{1}{2}\text{AD}$$[BF = AE]$
$\therefore$ $E$ is the mid-point of $AD$.
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