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

Solution:

In $\Delta AEP$ and $\triangle ADB$,

$\angle AEP =\angle ADB=90^o$

$\angle A=\angle A$             (common)

Therefore, by AA criterion,

$\Delta AEP \sim \Delta ADB$

Hence proved.

raja
Updated on 10-Oct-2022 13:21:08

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