$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB$. Show that:
$(i)$ $\Delta DAP \cong \Delta EBP$
$(ii)$ $AD = BE$
$(i)$ $\Delta DAP \cong \Delta EBP$
$(ii)$ $AD = BE$
(N/A) Given: $P$ is the mid-point of $AB$.
$\therefore AP = BP$
$\angle EPA = \angle DPB$ [Given]
Adding $\angle EPD$ to both sides,we get:
$\angle EPA + \angle EPD = \angle DPB + \angle EPD$
$\Rightarrow \angle APD = \angle BPE$
$(i)$ In $\Delta DAP$ and $\Delta EBP$:
$AP = BP$ [Proved above]
$\angle PAD = \angle PBE$ [Given as $\angle BAD = \angle ABE$]
$\angle APD = \angle BPE$ [Proved above]
By $ASA$ congruence criterion,we have:
$\Delta DAP \cong \Delta EBP$
$(ii)$ Since $\Delta DAP \cong \Delta EBP$,their corresponding parts are equal $(CPCT)$:
$\Rightarrow AD = BE$
$\therefore AP = BP$
$\angle EPA = \angle DPB$ [Given]
Adding $\angle EPD$ to both sides,we get:
$\angle EPA + \angle EPD = \angle DPB + \angle EPD$
$\Rightarrow \angle APD = \angle BPE$
$(i)$ In $\Delta DAP$ and $\Delta EBP$:
$AP = BP$ [Proved above]
$\angle PAD = \angle PBE$ [Given as $\angle BAD = \angle ABE$]
$\angle APD = \angle BPE$ [Proved above]
By $ASA$ congruence criterion,we have:
$\Delta DAP \cong \Delta EBP$
$(ii)$ Since $\Delta DAP \cong \Delta EBP$,their corresponding parts are equal $(CPCT)$:
$\Rightarrow AD = BE$
Explore More
Similar Questions
In the figure,sides $AB$ and $AC$ of $\Delta ABC$ are extended to points $P$ and $Q$ respectively. Also,$\angle PBC < \angle QCB$. Show that $AC > AB$.
Medium
View SolutionIn $\Delta ABC$,$AD$ is the perpendicular bisector of $BC$ (see figure). Show that $\Delta ABC$ is an isosceles triangle in which $AB = AC$.
Medium
View Solution$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$ (see figure). Show that $\angle ABD = \angle ACD$.
Medium
View Solution$\Delta ABC$ is an isosceles triangle in which altitudes $BE$ and $CF$ are drawn to equal sides $AC$ and $AB$ respectively. Show that these altitudes are equal.
Medium
View SolutionLine $l$ 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 $\angle A$.
Medium
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