In the figure,altitudes $AD$ and $CE$ of $\Delta ABC$ intersect each other at the point $P$. Show that $\Delta AEP \sim \Delta CDP$.
(N/A) In $\Delta AEP$ and $\Delta CDP$:
$\angle AEP = \angle CDP = 90^{\circ}$ (Since $AD \perp BC$ and $CE \perp AB$)
$\angle APE = \angle CPD$ (Vertically opposite angles)
Therefore,by the $AA$ similarity criterion,
$\Delta AEP \sim \Delta CDP$.
$\angle AEP = \angle CDP = 90^{\circ}$ (Since $AD \perp BC$ and $CE \perp AB$)
$\angle APE = \angle CPD$ (Vertically opposite angles)
Therefore,by the $AA$ similarity criterion,
$\Delta AEP \sim \Delta CDP$.
Explore More
Similar Questions
It is given that $\triangle FED \sim \triangle STU$. Is it true to say that $\frac{DE}{ST} = \frac{EF}{TU}$? Why?
Medium
View SolutionIn the figure,if $\angle ACB = \angle CDA$,$AC = 8 \, cm$ and $AD = 3 \, cm$,find $BD$ (in $cm$).
Difficult
View SolutionThe areas of two similar triangles are $144$ and $81$. If the length of the longest side of the first triangle is $36$,the length of the longest side of the second triangle is........
Medium
View SolutionIf $\triangle ABC \sim \triangle DEF$,$AB = 4 \, cm$,$DE = 6 \, cm$,$EF = 9 \, cm$,and $FD = 12 \, cm$,find the perimeter of $\triangle ABC$ (in $cm$).
Difficult
View SolutionIn $\Delta ABC$ and $\Delta PQR$,$m \angle A = m \angle R$ and $m \angle B = m \angle Q$. Then,the correspondence $ABC \leftrightarrow \ldots$ is a similarity.
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