In the given figure,if $\angle D = \angle C$,then is it true that $\triangle ADE \sim \triangle ACB$? Why?
(N/A) Yes,it is true.
In $\triangle ADE$ and $\triangle ACB$:
$1$. $\angle A = \angle A$ (Common angle in both triangles).
$2$. $\angle D = \angle C$ (Given in the problem).
Therefore,by the $AA$ similarity criterion,$\triangle ADE \sim \triangle ACB$.
In $\triangle ADE$ and $\triangle ACB$:
$1$. $\angle A = \angle A$ (Common angle in both triangles).
$2$. $\angle D = \angle C$ (Given in the problem).
Therefore,by the $AA$ similarity criterion,$\triangle ADE \sim \triangle ACB$.
Explore More
Similar Questions
The areas of two similar triangles are $25$ and $16$. Then,the ratio of their perimeters is $\ldots \ldots \ldots \ldots$
Medium
View SolutionIn $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4$ and $CM = 5$,find $AB$,$BC$,and $BM$.
Medium
View SolutionIn $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Then,$\angle BAM \cong \ldots$
Medium
View SolutionIn the figure,$D$ is a point on the hypotenuse $AC$ of $\Delta ABC,$ such that $BD \perp AC,$ $DM \perp BC,$ and $DN \perp AB.$ Prove that $DN^{2} = DM \cdot AN.$
Difficult
View SolutionIn parallelogram $ABCD$,$A-P-B$ and $AP = \frac{2}{3} AB$. $\overline{DP}$ intersects $\overline{AC}$ at $Q$. Find the ratio of the areas of $\Delta APQ$ and $\Delta CDQ$.
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