In ^@\triangle ABC^@, with ^@AB = AC^@, prove that the altitude | Math Question and Answer

Answer:

Step by Step Explanation:

We know that $△ A B C$ $△ A B C$ is an isosceles triangle in which $A B = A C$ $A B = A C$ .

Let $A D$ $A D$ be the altitude from the vertex $A$ $A$ on the side $B C$ $B C$ .

As the altitude from a vertex to the opposite side is perpendicular to the opposite side. $⟹ A D ⊥ B C$ Let us now represent the above situation with the help of a figure.
We need to prove that $BD = DC.$
In the right-angled $\triangle ADB$ and $\triangle ADC,$ we have $\begin{aligned} & AD = AD && [\text{Common}] \\ & AB = AC && [\text{Given}] \\ \therefore \space & \triangle ADB \cong \triangle ADC && [\text{By RHS criterion}] \end{aligned}$
As corresponding parts of congruent triangles are equal, we have $BD = DC$
Thus, in an isosceles triangle, the altitude from the vertex bisects the base.

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