Question

In Fig, 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) $△PDC\sim △BEC$

Answer 1

In $△AEP$ and $△CDP$,
$\angle APE=\angle CPD$ (Vertically opposite angle)
$\angle AEP=\angle CDP={90}^o$
$\therefore$ By AA criterion of similarity, $△AEP$ $\sim$ $△CDP$
In $△ABD$ and $△CBE$
$\angle ADB=\angle CEB={90}^o$
$\angle B$ is common
$\therefore$ By AA criterion of similarity, $△ABD$ $\sim$ $△CBE$
In $△AEP$ and $△ADB$
$\angle AEP=\angle ADB={90}^o$
$\angle A$ is common
$\therefore$ By AA criterion of similarity, $△AEP$ $\sim$ $△ADB$
$△PDC$ and $△BEC$
$\angle PDC=\angle BEC={90}^o$
$\angle C$ is common
$\therefore$ By AA criterion of similarity, $△PDC$ $\sim$ $△BEC$