In the figure, given below, the medians BD and CE of a triangle ABC meet at G. Prove that :

(i) △EGD ~ △CGB and

(ii) BG = 2GD from (i) above.

(i) Since, BD and CE are medians.

So, E is mid-point of AB and D is mid-point of AC.

By converse of mid-point theorem,

ED || BC and ED = $\dfrac{1}{2}$BC

$\dfrac{ED}{BC} = \dfrac{1}{2}$ …..(1)

In △EGD and △CGB,

∠EGD = ∠BGC (Vertically opposite angles are equal)

∠DEG = ∠GCB (Alternate angles are equal)

∴ △EGD ~ △CGB [By AA].

Hence, proved that △EGD ~ △CGB.

(ii) Since, corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{BG}{GD} = \dfrac{ED}{BC} \\[1em] \Rightarrow \dfrac{BG}{GD} = \dfrac{1}{2} \space …..\text{ (From 1)} \\[1em] \Rightarrow BG = 2GD.$

Hence, proved that BG = 2GD.