Question

$\Delta ABC$ is an isosceles $\Delta$ with $AB=AC$. If $BD$ and $CE$ are two medians of a triangle. Prove that $BD=CE$.

Answer 1

Let the sides of a $△ABC$ be $a,b,b$ because the triangle is isosceles

Median $BD=M_b=\frac{1}{2}\sqrt{2a^2+2b^2-b^2}=\frac{1}{2}\sqrt{2a^2+b^2}$ .......(1)

Median $CE=M_c=\frac{1}{2}\sqrt{2a^2+2b^2-b^2}=\frac{1}{2}\sqrt{2a^2+b^2}$ .....(2)

As per (1) and (2) we get

$\therefore M_b=M_c=\frac{1}{2}\sqrt{2a^2+b^2}$

Thus, $BD=CE$