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# In the given figure, altitudes $AD$ and $CE$ of $∆ABC$ intersect each other at the point $P$. Show that:(i) $∆AEP \sim ∆CDP$(ii) $∆ABD \sim ∆CBE$(iii) $∆AEP \sim ∆ADB$(iv) $∆PDC \sim ∆BEC$"

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

(i) $∆AEP \sim ∆CDP$

(ii) $∆ABD \sim ∆CBE$

(iii) $∆AEP \sim ∆ADB$

(iv) $∆PDC \sim ∆BEC$

Solution:

(i) In $\Delta AEP$ and $\triangle CDP$,

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

$\angle APE=\angle CPD$             (vertically opposite angles)

Therefore, by AA criterion,

$\Delta AEP \sim \Delta CDP$

Hence proved.

(ii) In $\Delta ABD$ and $\triangle CBE$,

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

$\angle ABD=\angle CBE$             (common)

Therefore, by AA criterion,

$\Delta ABD \sim \Delta CBE$

Hence proved.

(iii) 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.

(iv) 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.

Updated on: 10-Oct-2022

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