Question

Line $l$ is the bisector of an angle $\angle A$ and $B$ is any point on $l$. $BP$ and $BQ$ are perpendiculars from $B$ to the arms of $A$.

Show that:

1. $△APB\cong △AQB$
2. $BP=BQ$ or $B$ is equidistant from the arms of $A$

Answer 1

Step 1: Show that $△APB$ and $△AQB$ are congruent

It is given that the line $l$ is the bisector of an angle $\angle A$.

i.e., $\angle BAQ=\angle BAP$ $...\left(\mathrm{i}\right)$

Also, it is given that $BP$ and $BQ$ are perpendicular from $B$ to the arms of $A$.

i.e., $\angle AQB=\angle APB=90°$ $...\left(\mathrm{ii}\right)$

Now, in $△APB$ and $△AQB$,

$\angle AQB=\angle APB=90°$ [from equation $\left(\mathrm{i}\right)$]

$\angle BAQ=\angle BAP$ [from equation $\left(\mathrm{ii}\right)$]

$BA=BA$ [common side]

Then, by the AAS (Angle-Angle-Side) congruency criterion,

$△APB\cong △AQB$

Hence, it is proved that $△APB\cong △AQB$.

Step 2: Equate $BP$ and $BQ$

Now, as we know, the corresponding parts of the congruent triangles are congruent (CPCTC). So,

$\because$ $△APB\cong △AQB$

$\Rightarrow$ $BP=BQ$

i.e., the point $B$ is equidistant from the arms of the $\angle A$.

Hence, it is proved that $BP=BQ$.

Therefore, it is proved that

1. $\mathbf{△}\mathit{A}\mathit{P}\mathit{B}\mathbf{}\mathbf{\cong }\mathbf{}\mathbf{△}\mathit{A}\mathit{Q}\mathit{B}$
2. $\mathit{B}\mathit{P}\mathbf{}\mathbf{=}\mathbf{}\mathit{B}\mathit{Q}$ or $B$ is equidistant from the arms of $\mathbf{\angle }\mathit{A}$.