# Prove that if lengths of 2 medians of a triangle are equal then the triangle is isosceles ?

Jul 21, 2018

Given

In $\Delta A B C$, D and E are mid points of AB and AC.

Two medians $B E = C D$.

RTP: $\Delta A B C$ is isosceles.

Construction

$D , E$ are joined. $D X \mathmr{and} E Y$ are two perpendiculars drawn on BC.

Proof

$D \mathmr{and} E$ being mid points of $A B \mathmr{and} A C$ , the line $D E$ must be parallel to $B C$. Hence perpendiculars $D X = E Y$

Now in $\Delta C D X \mathmr{and} \Delta B E Y$

$\angle B Y E = \angle C X D = {90}^{\circ}$

$B E = C D$. given

and

$D X = E Y$ proved

$\Delta C D X \cong \Delta B E Y$ following $R H S$ rule.

So
$\angle E B Y = \angle D C X \mathmr{and} \angle E B C = \angle D C B$

Now in $\Delta D C B \mathmr{and} \Delta E B C$ we have

$B E = C D$.given

$B C$ common

and

$\angle E B C = \angle D C B$

So $\Delta D C B \cong \Delta E B C$ by SAS rule

So $\angle D B C = \angle E C B$

$\implies \angle A B C = \angle A C B$

This means $\Delta A B C$ is isosceles.