Question

$ABC$ is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal (see Fig.). Show that
(i) $△ABE\cong △ACF$
(ii) $AB=AC$, i.e., $ABC$ is an isosceles triangle.

Answer 1

To prove:

(i) $△ABE\cong △ACF$

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

Proof:

In $\Delta ABE$ and $\Delta ACF$,

$\angle BAE=\angle CAF$ (Common angle)

$\angle AEB=\angle AFC$ $....(\because BE\perp AC$ and $CF\perp AB)$

$BE=CF$ (Given that altitudes are equal)

By AAS criterion of congruence,

$\Delta ABE\cong \Delta ACF$ ---(1)

In $\Delta ABC$

$AB=AC$ (by CPCT)

Therefore, $\Delta ABC$ is an isosceles triangle. ---(2)

Hence, proved.