In the following figure, AD and CE are medians of ∆ABC. DF is drawn parallel to CE. Prove that:

(i) EF = FB,

(ii) AG : GD = 2 : 1

(i) In ∆BFD and ∆BEC,

∠BFD = ∠BEC [Corresponding angles are equal]

∠FBD = ∠EBC [Common]

∴ ∆BFD ~ ∆BEC [By AA].

Since, corresponding sides of similar triangles are proportional.

$\Rightarrow \dfrac{BF}{BE} = \dfrac{BD}{BC}$

$\Rightarrow \dfrac{BF}{BE} = \dfrac{1}{2}$ [∵ AD is median so D is the mid-point of BC]

$\Rightarrow BE = 2BF$

From figure,

⇒ BE = BF + FE

⇒ 2BF = BF + FE

⇒ BF = FE.

Hence, proved that EF = FB.

(ii) In ∆AFD, EG || FD.

By basic proportionality theorem we have,

$\dfrac{AE}{EF} = \dfrac{AG}{GD}$ …..(1)

Now, AE = EB [∵ CE is median so E is the mid-point of AB]

As, AE = EB = 2EF [As, EF = FB].

Substituting value of AE in (1) we get,

$\dfrac{AG}{GD} = \dfrac{2EF}{EF} = \dfrac{2}{1}$.

Hence, AG : GD = 2 : 1.