$ \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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