In the given figure, the median BD and CE of triangle ABC meet at point G. Show that BG = 2GD.

Since, BD and CE are medians.

∴ DA = DC and AE = BE.

∴ DE || BC (By converse of basic proportionality theorem)

In △EGD and △CGB,

⇒ ∠DEG = ∠GCB (Alternate angles are equal)

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

∴ △EGD ~ △CGB [By A.A. axiom]

We know that,

Ratio of corresponding sides of similar triangles are proportional.

⇒ $\dfrac{GD}{GB} = \dfrac{ED}{BC}$ …………..(1)

In △AED and △ABC,

⇒ ∠AED = ∠ABC (Corresponding angles are equal)

⇒ ∠EAD = ∠BAC (Common)

∴ △AED ~ △ABC [By A.A. axiom]

⇒ $\dfrac{ED}{BC} = \dfrac{AE}{AB}$ …………..(2)

As, E is the mid-point of AB. So,

⇒ $\dfrac{AE}{AB} = \dfrac{1}{2}$

Substituting value of $\dfrac{AE}{AB}$ in equation (2), we get :

⇒ $\dfrac{ED}{BC} = \dfrac{1}{2}$.

Substituting value of $\dfrac{ED}{BC}$ in equation (1), we get :

⇒ $\dfrac{GD}{GB} = \dfrac{1}{2}$

⇒ GB = 2GD.

Hence, proved that GB = 2GD.