The median to the base of an isosceles triangle is perpendicular to the base

• In isosceles and equilateral triangles, the median drawn from the vertex bisects the angle whose two adjacent sides are equal.
• The median not only bisects the side opposite to the vertex, but it also bisects the angle of the vertex in the case of equilateral and isosceles triangles.

An isosceles triangle with median on base is shown below,

Here, side$\mathit{A}$ Aand$\mathit{B}$is equal. Thus,$\left|A\right|$ is equal to$\left|B\right|$

To write vector$C$ in terms of$A$ and$B$ as follows:

$C=B-A$

Write the expression of median in vector addition form as follows:

$M=A+\frac{C}{2}$

Substitute $(B-A)$for $C$in above equation as follows:

$M=A+\frac{(B-A)}{2}$

$M=\frac{1}{2}(B+A)$

The dot product of base and median vector is calculated as follows:

$$\begin{gathered} M\times C=\frac{1}{2}(B+A)\times (B-A) \\ M\times C=\frac{1}{2}\left(|A{|}^2-|B{|}^2\right)\left(\right|A|=|B\left|\right) \\ M\times C=\frac{1}{2}\left(|A{|}^2-|A{|}^2\right) \\ M\times C=0 \end{gathered}$$

Therefore, the median of an isosceles triangle is orthogonal to the base.