### If the altitude from one vertex of a triangle bisects the opposite side, prove that the triangle is isosceles.

Answer:

Step by Step Explanation:
1. We know that an altitude from a vertex to the opposite side is the perpendicular drawn from that vertex to the opposite side.

Let us now draw the altitude $AD$ from the vertex $A$ of $\triangle ABC$ to the opposite side $BC$.
As the altitude $AD$ bisects the opposite side $BC$, $$BD = DC$$.
2. We need to prove that the triangle is isosceles, i.e., $AB = AC$.
3. In $\triangle ADB$ and $\triangle ADC$, we have \begin{aligned} BD = DC && \text { [Given] } \\ AD = AD && \text{ [Common] } \\ \angle ADB = \angle ADC = 90^ \circ && \text{[As } AD \perp BC \text{]}\\ \therefore \triangle ADB \cong \triangle ADC && \text{ [By SAS congruence criterion] } \\ \end{aligned}
4. As the corresponding parts of congruent triangles are equal, we have $AB = AC$.
Thus, $\triangle ABC$ is an isosceles triangle.

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