If the altitudes from the two vertices of a triangle to the oppo | Math Question and Answer

Answer:

Step by Step Explanation:

Let $B D$ and $C E$ be the altitudes from the vertices $B$ and $D$ of $△ A B C .$

As the altitude of a triangle is perpendicular to the opposite side, we have $B D ⊥ A C and C E ⊥ A B$ Also, we are told that the altitudes are of equal length. $⟹ B D = C E$
We need to prove that $AB = AC.$
In $\triangle ADB$ and $\triangle AEC$ , we have $\begin{aligned} &BD = CE &&[\text{Given}] \\ &\angle BAD = \angle CAE &&[\text{Common}]\\ &\angle ADB = \angle AEC = 90^ \circ &&[ BD \perp AC \text { and } CE \perp AB] \\ \therefore \space &\triangle ADB \cong \triangle AEC &&[\text{By AAS criterion}] \end{aligned}$
As the corresponding parts of congruent triangles are equal, we have $AB = AC .$
Hence, $\triangle ABC$ is isosceles.

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