In the figure,altitudes $AD$ and $CE$ of $\Delta ABC$ intersect each other at the point $P$. Show that $\Delta PDC \sim \Delta BEC$.
(N/A) In $\Delta PDC$ and $\Delta BEC$:
$\angle PDC = \angle BEC = 90^{\circ}$ (Since $AD \perp BC$ and $CE \perp AB$)
$\angle PCD = \angle BCE$ (Common angle)
Therefore,by using the $AA$ similarity criterion,
$\Delta PDC \sim \Delta BEC$.
$\angle PDC = \angle BEC = 90^{\circ}$ (Since $AD \perp BC$ and $CE \perp AB$)
$\angle PCD = \angle BCE$ (Common angle)
Therefore,by using the $AA$ similarity criterion,
$\Delta PDC \sim \Delta BEC$.
Explore More
Similar Questions
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle,write the length of its hypotenuse.
$50 \text{ cm}, 80 \text{ cm}, 100 \text{ cm}$
$50 \text{ cm}, 80 \text{ cm}, 100 \text{ cm}$
Medium
View SolutionIn the figure,$A, B$ and $C$ are points on $OP, OQ$ and $OR$ respectively such that $AB \parallel PQ$ and $AC \parallel PR$. Show that $BC \parallel QR$.
Difficult
View Solution$S$ and $T$ are points on sides $PR$ and $QR$ of $\Delta PQR$ such that $\angle P = \angle RTS$. Show that $\Delta RPQ \sim \Delta RTS$.
Difficult
View SolutionIn the figure,if $LM \parallel CB$ and $LN \parallel CD$,prove that $\frac{AM}{AB} = \frac{AN}{AD}$.
Medium
View SolutionIn the figure,$ABC$ is a triangle in which $\angle ABC < 90^{\circ}$ and $AD \perp BC$. Prove that $AC^{2} = AB^{2} + BC^{2} - 2BC \cdot BD$.
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