Question

In fig., altitude \[AD\] and \[CE\] of \[\Delta ABC\] intersect each other at the point \[P\].
Show that:
A) \[\Delta AEP \sim \Delta CDP\]
B) \[\Delta ABD \sim \Delta CBE\]
C) \[\Delta AEP \sim \Delta ADB\]
D) \[\Delta PDC \sim \Delta BEC\]

Answer 1

Hint: We know that the altitudes of triangles form right angles.
We also know that; the vertically opposite angles are equal to each other.
Using these conditions and diagrams we can find the condition of similarity of the triangles.
For any two right-angle triangles, if one of the other two angles is equal then the third angle is also equal to each other. This condition is known as A-A-A similarity condition. Then, we can say that the triangles are similar.

Complete step-by-step answer:
It is given that; \[AD\] and \[CE\] are the altitudes of \[\Delta ABC\] intersect each other at the point \[P\]
We have to show that,
\[\Delta AEP \sim \Delta CDP\]
\[\Delta ABD \sim \Delta CBE\]
\[\Delta AEP \sim \Delta ADB\]
\[\Delta PDC \sim \Delta BEC\]
To prove: \[\Delta AEP \sim \Delta CDP\]
Since, \[AD\] and \[CE\] are the altitudes of \[\Delta ABC\].
\[\angle AEP = \angle CDP\], as each angle is \[{90^ \circ }\].
In \[\Delta AEP\] and \[\Delta CDP\]
\[\angle AEP = \angle CDP\]
\[\angle APE = \angle CPD\] as they are vertically opposite angles.
So, by A-A-A similarity \[\Delta AEP \sim \Delta CDP\].
To prove: \[\Delta ABD \sim \Delta CBE\]
Since, \[AD\] and \[CE\] are the altitudes of \[\Delta ABC\].
\[\angle ADB = \angle CEB\], as each angle is \[{90^ \circ }\].
In \[\Delta ABD\] and \[\Delta CBE\]
\[\angle ADB = \angle CEB\]
\[\angle ABD = \angle CBE\] as they are common angles.
So, by A-A-A similarity \[\Delta ABD \sim \Delta CBE\].
To prove: \[\Delta AEP \sim \Delta ADB\]
Since, \[AD\] and \[CE\] are the altitudes of \[\Delta ABC\].
\[\angle AEP = \angle ADB\], as each angle is \[{90^ \circ }\].
In \[\Delta ABD\] and \[\Delta CBE\]
\[\angle AEP = \angle ADB\]
\[\angle PAE = \angle DAB\] as they are common angles.
So, by A-A-A similarity \[\Delta AEP \sim \Delta ADB\].
To prove: \[\Delta PDC \sim \Delta BEC\]
Since, \[AD\] and \[CE\] are the altitudes of \[\Delta ABC\].
\[\angle CEB = \angle CDP\], as each angle is \[{90^ \circ }\].
In \[\Delta ABD\] and \[\Delta CBE\]
\[\angle CEB = \angle CDP\]
\[\angle BCE = \angle PCD\] as they are common angles.
So, by A-A-A similarity \[\Delta PDC \sim \Delta BEC\].
Hence,
\[\Delta AEP \sim \Delta CDP\]
\[\Delta ABD \sim \Delta CBE\]
\[\Delta AEP \sim \Delta ADB\]
\[\Delta PDC \sim \Delta BEC\]

Note: If the three angles of any triangle are equal to the respective angles of another triangle, then the triangles are called similar triangles.
The altitude (also known as the height) of a triangle is perpendicular to the base.
When two lines intersect each other, then the opposite angles formed due to intersection are called vertical angles or vertically opposite angles. A pair of vertically opposite angles are always equal to each other.