In the figure (2) given below, medians BE and CF of a △ABC meet at G. Prove that :

(i) △FGE ~ △CGB

(ii) BG = 2GE

(i) Considering △FGE and △CGB,

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

∠GFE = ∠GCB (Alternate angles are equal)

Hence by AA axiom △FGE ~ △CGB.

(ii) Considering △AFE and △ABC,

∠A = ∠A (Common angles)

∠AFE = ∠ABC (Corresponding angles are equal)

Hence by AA axiom △AFE ~ △ABC.

Given BE is the median of AC, so

AE = EC

AC = AE + EC = AE + AE = 2AE.

$\therefore \dfrac{AE}{AC} = \dfrac{1}{2}.$

Since, △AFE ~ △ABC, so the ratio of their corresponding sides are equal,

$\therefore \dfrac{FE}{BC} = \dfrac{AE}{AC} \\[1em] \Rightarrow \dfrac{FE}{BC} = \dfrac{1}{2}.$

Since, △FGE ~ △CGB, so the ratio of their corresponding sides are equal,

$\therefore \dfrac{FE}{BC} = \dfrac{GE}{BG} \\[1em] \Rightarrow \dfrac{GE}{BG} = \dfrac{1}{2} \\[1em] \Rightarrow BG = 2GE.$

Hence, proved that BG = 2GE.