In the figure,if $\Delta ABE \cong \Delta ACD,$ show that $\Delta ADE \sim \Delta ABC.$
(N/A) Given: $\Delta ABE \cong \Delta ACD.$
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$ $...(1)$
And,$AE = AD$ $...(2)$
Now,consider $\Delta ADE$ and $\Delta ABC$:
From equations $(1)$ and $(2)$,we have:
$\frac{AD}{AB} = \frac{AE}{AC}$
Also,$\angle DAE = \angle BAC$ (Common angle).
Therefore,by the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ADE \sim \Delta ABC.$
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$ $...(1)$
And,$AE = AD$ $...(2)$
Now,consider $\Delta ADE$ and $\Delta ABC$:
From equations $(1)$ and $(2)$,we have:
$\frac{AD}{AB} = \frac{AE}{AC}$
Also,$\angle DAE = \angle BAC$ (Common angle).
Therefore,by the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ADE \sim \Delta ABC.$
Explore More
Similar Questions
In the figure,$DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$.
Medium
View SolutionIn the figure,$\angle ACB = 90^{\circ}$ and $CD \perp AB$. Prove that $\frac{BC^{2}}{AC^{2}} = \frac{BD}{AD}$.
Medium
View SolutionIn the figure,$AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that $AC^2 + AB^2 = 2AD^2 + \frac{1}{2} BC^2$.
Difficult
View SolutionIn the figure,$\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\Delta PQS \sim \Delta TQR$.
Medium
View SolutionIn the figure,altitudes $AD$ and $CE$ of $\Delta ABC$ intersect each other at the point $P$. Show that $\Delta AEP \sim \Delta ADB$.
Easy
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