Question
In $\triangle ABC,$ altitudes $\text{BE}$ and $\text{CF}$ are equal. Prove that the triangle is isosceles.
✓
Answer
In $\triangle ABE$ and $\triangle ACF$,
$\angle A=\angle A ~ \ldots . . . . . .[$Common$]$
$\angle AEB =\angle AFC =90^{\circ} \ldots . .[$ Given: $\text{BE} \perp \text{AC} ; \text{CF} \perp \text{AB} ]$
$\text{BE} = \text{CF} ..........[$Given$]$
$\therefore \triangle ABE \cong \triangle ACE \ldots . . . . .[ \text {AAS}$ Criterion]
$ \Rightarrow \text{AB} = \text{AC}$
Therefore, $\text{ABC}$ is an isosceles triangle.
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