Question

$ABC$ is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal (see the given figure). Show that

(i) $△ABE\cong △ACF$

(ii) $AB=AC,$ i.e., $ABC$ is an isosceles triangle.

Answer 1

(i) In $△ABE$ and $△ACF,$ we have

$\angle AEB=\angle AFC$ (Each ${90}^{\circ }$)

$\angle A=\angle A$ (Common angle)

$BE=CF$ (Given)

$\therefore △ABE\cong △ACF$ (By AAS congruence rule)

(ii) It has already been proved that

$△ABE\cong △ACF$

$\Rightarrow AB=AC$ (By CPCT)

So, $ABC$ is an isosceles triangle.