If the medians of a intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = ar (ABC)

If the medians of a ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = 1/3 ar (ABC)

If the medians of a $\triangle ABC$ intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = $\frac{1}{3}$ ar (ABC)

Answers (1)

Solution.

Given: $\triangle ABC$ with medians AM, BN & CL
$ar (AGB) = ar (AGC) = ar (BGC) =\frac{1}{3} ar (ABC)$

Proof: We know that a median divides the triangle into two triangles of same area.
Let the area of small triangles be denoted as 1, 2, 3, 4, 5, 6 as shown in the figure
AM is the median
$ar (\triangle AMC) = ar (\triangle AMB)$
$ar(1) + ar(2) + ar(6) = ar(3) + ar(4) + ar(5)$   …(i)
CL is the median
$ar(\triangle CAL)=ar(\triangle CBL)$
$ar(1)+ar(6)+ar(5)=ar(2)+ar(3)+ar(4)$   …(ii)
BN is the median
$ar(\triangle BNA)=ar(\triangle BNC)$
$ar(4)+ar(5)+ar(6)=ar(1)+ar(2)+ar(3)$
(i) – (ii)
$ar(1)+ar(2)+ar(6)-ar(1)-ar(6)-ar(5)=ar(3)+ar(4)+ar(5)-ar(2)-ar(3)-ar(4)$

So, $ar(2)=ar(5)$
(ii) – (iii)
$ar(1)+ar(6)+ar(5)-ar(4)-ar(5)-ar(6)=ar(2)+ar(3)+ar(4)-ar(1)-ar(2)-ar(3)$
$ar(1)-ar(4)=ar(4)-ar(1)$
So, $ar(1)=ar(4)$
Similarly we can prove that area of $ar(1)=ar(2)=ar(3)=ar(4)=ar(5)=ar(6)$
Hence the triangle is divided into 6 triangles of equal area.
$\triangle AGB, \triangle AGC$ and $\triangle BGC$ consists of 2 triangles each
So they have equal area which is equal to twice the area of the smaller triangles.
$ar(\triangle AGB)=ar(\triangle AGC)=ar(\triangle BGC)=2$    (area of 1 small triangle)
Hence, $ar(\triangle AGB)=ar(\triangle AGC)=ar(\triangle BGC)=\frac{1}{3}ar(ABC)$
Hence proved.