# $\mathrm{ABC}$ is a triangle in which altitudes $\mathrm{BE}$ and $\mathrm{CF}$ to sides $\mathrm{AC}$ and $\mathrm{AB}$ are equal (see Fig. 7.32). Show that(i) $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACF}$(ii) $\mathrm{AB}=\mathrm{AC}$, i.e., $\mathrm{ABC}$ is an isosceles triangle."

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Given:

$ABC$ is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal.

To do:

We have to show that

(i) $\triangle ABE \cong \triangle ACF$

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

Solution:

(i) We know that,

If two angles of a triangle with a non-included side are equal to the corresponding angles and non-included side of the other triangle, they are considered to be congruent.

Given, $BE=CF$

We have,

$\angle A$ as the common angle of $AEB$ and $AFC$

$\angle AEB=\angle AFC$

Therefore,

$\triangle AEB \cong \triangle AFC$

We also know

From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding  sides must be equal.

Therefore,

$AB=AC$.

Hence, $ABC$ is an isosceles triangle.

Updated on 10-Oct-2022 13:41:10