If the altitude from one vertex of a triangle bisects the opposi | Math Question and Answer

Answer:

Step by Step Explanation:

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

You can reuse this answer
Creative Commons License

Chat with us