In the adjoining figure, the medians BD and CE of a ΔABC meet at G. Prove that :

(i) ΔEGD ∼ ΔCGB.

(ii) BG = 2 × GD.

(i) Since, BD and CE are medians.

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

By mid-point theorem,

ED ∥ BC and ED = $\dfrac{1}{2}$ BC [By mid-point theorem]

$\Rightarrow \dfrac{ED}{BC} = \dfrac{1}{2}$

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

In triangle EGD and BGC,

∠EGD = ∠BGC [Vertically opposite angles are equal]

∠DEG = ∠GCB [Alternate angles are equal]

∴ ΔEGD ∼ ΔCGB by (By A.A. axiom)

Hence, proved that ΔEGD ∼ ΔCGB.

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

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

Hence, proved that BG = 2GD.