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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.
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