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

Given, altitudes AD and CE of \(\triangle\) ABC intersect each other at the point P.
(i) In \(\triangle\) AEP and \(\triangle\) CDP,
\(\angle\) AEP = \(\angle\) CDP (90° each)
\(\angle\) APE = \(\angle\) CPD (Vertically opposite angles)
Hence, by AA similarity criterion,
\(\triangle\) AEP ~ \(\triangle\) CDP

(ii) In \(\triangle\) ABD and \(\triangle\) CBE,
\(\angle\) ADB = \(\angle\) CEB ( 90° each)
\(\angle\) ABD = \(\angle\) CBE (Common Angles)
Hence, by AA similarity criterion,
\(\triangle\) ABD ~ \(\triangle\) CBE

(iii) In \(\triangle\) AEP and \(\triangle\) ADB,
\(\angle\) AEP = \(\angle\) ADB (90° each)
\(\angle\) PAE = \(\angle\) DAB (Common Angles)
Hence, by AA similarity criterion,
\(\triangle\) AEP ~ \(\triangle\) ADB

(iv) In \(\triangle\) PDC and \(\triangle\) BEC,
\(\angle\) PDC = \(\angle\) BEC (90° each)
\(\angle\) PCD = \(\angle\) BCE (Common angles)
Hence, by AA similarity criterion,
\(\triangle\) PDC ~ \(\triangle\) BEC
Hence, by AA similarity criterion,
\(\triangle\) PDC ~ \(\triangle\) BEC