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 $\angle A$ (see Fig). Show that :
$(i)$ $\Delta APB \cong \Delta AQB$
$(ii)$ $BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.
$(i)$ $\Delta APB \cong \Delta AQB$
$(ii)$ $BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.
(N/A) Given: Line $l$ is the bisector of $\angle A$. $BP \perp AQ$ and $BQ \perp AP$.
$(i)$ In $\Delta APB$ and $\Delta AQB$:
$\angle APB = \angle AQB = 90^{\circ}$ (Given)
$\angle PAB = \angle QAB$ ($l$ is the bisector of $\angle A$)
$AB = AB$ (Common side)
Therefore,by $AAS$ congruence criterion,$\Delta APB \cong \Delta AQB$.
$(ii)$ Since $\Delta APB \cong \Delta AQB$,their corresponding parts are equal by $CPCT$.
Therefore,$BP = BQ$.
This shows that point $B$ is equidistant from the arms of $\angle A$.
$(i)$ In $\Delta APB$ and $\Delta AQB$:
$\angle APB = \angle AQB = 90^{\circ}$ (Given)
$\angle PAB = \angle QAB$ ($l$ is the bisector of $\angle A$)
$AB = AB$ (Common side)
Therefore,by $AAS$ congruence criterion,$\Delta APB \cong \Delta AQB$.
$(ii)$ Since $\Delta APB \cong \Delta AQB$,their corresponding parts are equal by $CPCT$.
Therefore,$BP = BQ$.
This shows that point $B$ is equidistant from the arms of $\angle A$.
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