$ \mathrm{ABC} $ is an isosceles triangle in which altitudes $ \mathrm{BE} $ and $ \mathrm{CF} $ are drawn to equal sides $ \mathrm{AC} $ and $ \mathrm{AB} $ respectively (see Fig. 7.31). Show that these altitudes are equal.
"

Given:

$ABC$ is an isosceles triangle in which altitudes $BE$ and $CF$ are drawn to equal sides $AC$ and $AB$ respectively.

To do:

We have to show that these altitudes are equal.

Solution:

Let us consider $\triangle AEB$ and $\triangle AFC$

We know that,

From Angle-Angle-Side:

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.

$AB=AC$ (given)

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,

$BE=CF$.

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Updated on: 10-Oct-2022

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