In the given figure, altitudes $AD$ and $CE$ of $∆ABC$ intersect each other at the point $P$. Show that $∆PDC \sim ∆BEC$
"

AcademicMathematicsNCERTClass 10

Given:

In the given figure, altitudes $AD$ and $CE$ of $∆ABC$ intersect each other at the point $P$.

To do:

We have to show that $∆PDC \sim ∆BEC$

Solution:

In $\Delta PDC$ and $\triangle BEC$,

$\angle PDC =\angle BEC=90^o$

$\angle PCD=\angle BCE$             (common)

Therefore, by AA criterion,

$\Delta PDC \sim \Delta BEC$

Hence proved.

raja
Updated on 10-Oct-2022 13:21:09

Advertisements