In the given figure, medians AD and BE of ΔABC meet at G and DF ∥ BE. Prove that :

(i) EF = FC.

(ii) AG : GD = 2 : 1.

(i) In ∆CFD and ∆CEB,

∠CDF = ∠CBE [Corresponding angles are equal]

∠FCD = ∠ECB [Common]

∴ ∆CFD ∼ ∆CEB (By A.A. axiom)

Since, corresponding sides of similar triangles are proportional.

$\Rightarrow \dfrac{CF}{CE} = \dfrac{CD}{CB} \\[1em] \Rightarrow \dfrac{CF}{CE} = \dfrac{1}{2} \text{ [∵ D is the mid-point of BC]} \Rightarrow CE = 2CF.$

From figure,

⇒ CE = CF + FE

⇒ 2CF = CF + FE

⇒ CF = FE.

Hence, proved that EF = FC.

(ii) In ∆AFD,

GE ∥ DF

By basic proportionality theorem we have,

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

Now, AE = EC [∵ BE is media,n so E is the mid-point of AC]

As, AE = EC = 2EF [As, EF = FC].

Substituting value of AE in (1) we get,

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

Hence, proved that AG : GD = 2 : 1.