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

Given:

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

To do:

We have to show that

(i) $âˆ†AEP \sim âˆ†CDP$

(ii) $âˆ†ABD \sim âˆ†CBE$

(iii) $âˆ†AEP \sim âˆ†ADB$

(iv) $âˆ†PDC \sim âˆ†BEC$

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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