Question

In the figure, if \( BD \perp AC \) and \( CE \perp AB \), prove that \( \triangle AEC \sim \triangle ADB \).

Hint:

For proving similarity of triangles, use the AA criterion, which states that two triangles are similar if two corresponding angles are equal.

Answer 1

We are given that: - \( BD \perp AC \), i.e., \( BD \) is perpendicular to \( AC \),
- \( CE \perp AB \), i.e., \( CE \) is perpendicular to \( AB \),
and we are required to prove that \( \triangle AEC \sim \triangle ADB \).
Step 1: Identify the right angles.
Since \( BD \perp AC \), we have \( \angle ADB = 90^\circ \).
Also, since \( CE \perp AB \), we have \( \angle AEC = 90^\circ \).

Step 2: Use the AA (Angle-Angle) similarity criterion.
To prove that \( \triangle AEC \sim \triangle ADB \), we need to show that two corresponding angles are equal. 1. We already know that \( \angle ADB = \angle AEC = 90^\circ \).
2. \( \angle BAD \) and \( \angle EAC \) are common angles for both triangles \( \triangle AEC \) and \( \triangle ADB \).
Since two corresponding angles in \( \triangle AEC \) and \( \triangle ADB \) are equal, by the AA similarity criterion, we can conclude that: \[ \triangle AEC \sim \triangle ADB. \]
Conclusion:
We have proved that \( \triangle AEC \sim \triangle ADB \) by the AA similarity criterion.