In the figure, $ABC$ is a triangle in which altitudes $BE$ and $CF$to sides $AC$ and $AB$ are equal. Show that

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

Hint: We will use the given conditions, that $BE = CF$ , \[\angle AEB = {90^ \circ }\] and \[\angle AFC = {90^ \circ }\]to prove the triangles, $\vartriangle ABC$ and $\vartriangle ACF$congruent. The triangles will be congruent by AAS property. Then, we will use the CPCT property to prove that $AB = AC$ or $ABC$ is an isosceles triangle.

Complete step-by-step answer:
We are given that altitudes $BE$ and $CF$to sides $AC$ and $AB$ are equal
So BE=CF
Hence, we have \[\angle AEB = {90^ \circ }\] and \[\angle AFC = {90^ \circ }\]
We have to prove that $\vartriangle ABC \cong \vartriangle ACF$
Consider $\vartriangle ABC$ and $\vartriangle ACF$.
We have already seen that \[\angle AEB = \angle AFC = {90^ \circ }\]
Here, $\angle A$ is common in both the triangles.
And $BE = CF$ is given
Therefore, $\vartriangle ABC \cong \vartriangle ACF$ by AAS rule of congruence.
Since, $\vartriangle ABC \cong \vartriangle ACF$, their corresponding parts will also be equal.
Therefore, we will have $AB = AC$.
And if in a triangle, two sides are equal, then we call it an isosceles triangle.

Note: Two congruent triangles have all corresponding sides and angles equal. Students must know various properties to prove two triangles congruent. Also, if two sides are equal, then it is an isosceles triangle and if all sides are equal, then we call it an equilateral triangle.