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 $\angle A$.
Explore More
Similar Questions
In an isosceles triangle $ABC$ with $AB = AC$,$D$ and $E$ are points on $BC$ such that $BE = CD$ (see Fig). Show that $AD = AE$.
Medium
View Solution$AD$ and $BC$ are equal perpendiculars to a line segment $AB$ (see figure). Show that $CD$ bisects $AB$.
Medium
View SolutionIn the figure,$OA = OB$ and $OD = OC$. Show that
$(i)$ $\Delta AOD \cong \Delta BOC$ and
$(ii)$ $AD \parallel BC$.
$(i)$ $\Delta AOD \cong \Delta BOC$ and
$(ii)$ $AD \parallel BC$.
Medium
View Solution$AB$ is a line segment. $P$ and $Q$ are points on opposite sides of $AB$ such that each of them is equidistant from the points $A$ and $B$ (see figure). Show that the line $PQ$ is the perpendicular bisector of $AB$.
Difficult
View Solution$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig). If $AD$ is extended to intersect $BC$ at $P$,show that:
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo