Question

In the given figure, D and G are mid points of side BC and median AD of triangle ABC . Prove that area of triangle ABC = 4ar(triangle BGD)

Answer 1

given =>
 AD is the median,i.e,BD=DC
andG is the M.P of AD
to prove=>
Ar. of tri.ABC=4 BGD
PROOF
Ar. of ABD=1/2 Ar. of tri. ABC  [by mid point theorem]
now,as G is the M.P of side AD of tri.ABD, BG is the median
so, Ar. of tri.BGD= 1/2 Ar. ofADB
It's proved before that Ar.of ADB is 1/2 Ar.ABC
so,
Ar.of bgd=1/2*[1/2 Ar. of ABC]= 1/4 Ar.of ABC
in other words,
Ar.of ABC= 4 Ar.of BGD
HENCE THE PROOF :)

Answer 2

Given =  AD is the median of Δ ABC.
To prove = ar BGD = ar 4 ABC
 PROOF =So , ar ( Δ BAD ) = ar (Δ ADC)  ( a median divides a triangle into two triangles of equal areas) - 1 

As, G is the midpoint of AD... so BG is the median.
so, ar (ΔBGD) = ar (Δ BGA) ( a median divides a triangle into two triangles of equal areas) -2 

and, CG is also the median because it is joining the mid point of AD i,e G

ar (Δ DGC) = ar (ΔAGC) ( a median divides a triangle into two triangles of equal areas) -3 

so, acc. to eqn 1, 2 ,3
ar (Δ BGD) = ar (ΔAGC)

 so ABC = 4ar(triangle BGD)