Question

In an acute angled triangle $ABC$, the internal bisector of angle $A$ meets base $BC$ at point $D$. $DE\perp AB$ and $DF\perp AC$; then the traingle $AEF$ is an isosceles triangle

• A. True
• B. False

Answer 1

The correct option is A True

Given: AD bisects $\angle A$, $DE\perp AB$ and $DF\perp AC$
To prove: AEF is an isosceles triangle
In $△AED$ and $△AFD$,
$AD=AD$ (Common)
$\angle AED=\angle AFD$ (Each ${90}^{\circ }$)
$\angle EAD=\angle FAD$ (AD bisects $\angle A$)
Thus, $△AED\cong △AFD$ (ASA rule)
Hence, $AE=AF$ (By cpct)
thus, $△AEF$ is an Isosceles triangle.