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# $ \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.

"

**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.

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