# 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)

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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

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)