Question

In $△$ ABC, altitudes BE and CF are equal, therefore the triangle is isosceles.
State true or false.

Answer 1

In $△ABE$ and $△ACF$
$\angle BAE=\angle CAF$ (Common angle)
$\angle AFB=\angle AFC$ (each ${90}^{\circ }$)
$BE=CF$ (Given)
Thus, $△ABF\cong △ACE$ (ASA rule)
Hence, $AB=AC$ (By cpct)
Or ABC is an Isosceles triangle