Question

In the given figure, AD is an altitude of the triangle $△ABC.$ Also, $BD=DE$. Find the measure of the angle $\angle p.$

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. Insufficient data.

Answer 1

The correct option is B ${45}^o$

As AD is an altitude of the triangle AC, it is perpendicular to the side BC.
$\therefore \angle BDA={90}^o$

Also, $\angle BDE=\angle BDA$
$\therefore \angle BDE={90}^o$

The sides BD and DE are equal in measure.
$\therefore △BDE$ is isosceles.
Hence, $\angle DBE=\angle BED=\angle p$

Also,sum of interior angles of a triangle is ${180}^o.$
$\angle DBE+\angle BED+\angle BDE={180}^o$
$\Rightarrow \angle p+\angle p+{90}^o={180}^o$
$\Rightarrow 2\angle p={90}^o$
$\Rightarrow \angle p=\frac{{90}^o}{2}={45}^o$
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