Question:
ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE.
Solution:
Given ΔABC is an isosceles triangle in which AB = AC and BD, CE are its two medians.
To show BD = CE.
Proof in $\triangle A B D$ and $\triangle A C E$, $A B=A C$ [given]
$\angle A=\angle A$ [common angle]
and $A D=A E$
$\because$ $A B=A C$
$\Rightarrow$ $\frac{1}{2} A B=\frac{1}{2} A C$
$\Rightarrow$ $A E=A D$
As $D$ is the mid-point of $A C$ and $E$ is the mid-point of $A B$.
$\therefore$ $\triangle A B D \cong \triangle A C E$ [by SAS congruence rule]
$\Rightarrow$ $B D=C E$ [by CPCT]