Given that \(AD\) and \(CE\) are altitudes of \(\triangle ABC\) and meet at \(P\), establish that:
(i) \(\Delta AEP \sim \Delta CDP\)
(ii) \(\Delta ABD \sim \Delta CBE\)
(iii) \(\Delta AEP \sim \Delta ADB\)
(iv) \(\Delta PDC \sim \Delta BEC\)
(i) In \(\Delta ABC\) altitude \(AD\) and \(CE\) of triangle intersects each other at the point \(P\).
In \(\Delta AEP\) and \(\Delta CDP\),
\(\angle AEP \ = \) (Since \(AD\) and \(CE\) are altitudes)
\(\angle APE\ =\) (Vertically opposite angles)
Thus, \(\Delta AEP \sim\) by similarity
Hence proved.
(ii) In \(\Delta ABD\) and \(\Delta CBE\)
\(\angle ABD =\) (common angle)
\(\angle ADB =\) (since \(AD\) and \(CE\) are altitudes, both \(90^\circ\))
Thus, \(\Delta ABD \sim\) (by similarity)
Hence proved.
(iii) In \(\Delta AEP\) and \(\Delta ADB\)
\(\angle PAE = \) (common angle)
\(\angle AEP = \) (since \(AD\) and \(CE\) are altitudes, both \(90^\circ\))
Thus, \(\Delta AEP \sim\) (by similarity)
Hence proved.
(iv) In \(\Delta BEC\) and \(\Delta PDC\)
\(\angle BCE = \) (common angle)
\(\angle CEB = \) (since \(AD\) and \(CE\) are altitudes, both \(90^\circ\))
Thus, \(\Delta BEC \sim \) ( by similarity)
Hence proved.