Question

In an acute-angled triangle ABC, the internal bisector of angle A meets base BC at point D. DE $\perp$ AB and DF $\perp$ AC; hence AEF is an isosceles triangle.
State whether the above statement is true or false.

• A. True
• B. False

Answer 1

The correct option is A True

In $△ADE$ and $△ADF$,
$\angle DAE=\angle DAF$ (AD bisects $\angle A$)
$AD=AD$ (Common)
$\angle AED=\angle AFD$ (Each ${90}^{\circ }$)
Thus, $△ADE\cong △ADF$ (ASA postulate)
Thus, AE = AF (By cpct)
Hence, $△AFE$ is an isosceles triangle