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  • 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 $\triangle A B C$ intersect at $G$, show that $\operatorname{ar}(\mathrm{AGB})=\operatorname{ar}(\mathrm{AGC})=\operatorname{ar}(\mathrm{BGC})=\frac{1}{3}$ ar $(\mathrm{ABC})$

Answers (1)

Solution.

Given: $\triangle A B C$ with medians AM, $\mathrm{BN} \& \mathrm{CL}$

$
\operatorname{ar}(A G B)=\operatorname{ar}(A G C)=\operatorname{ar}(B G C)=\frac{1}{3} \operatorname{ar}(A B C)
$
 


Proof:

We know that a median divides the triangle into two triangles of the 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

$
\begin{aligned}
& \operatorname{ar}(\triangle A M C)=\operatorname{ar}(\triangle A M B) \\
& \operatorname{ar}(1)+\operatorname{ar}(2)+\operatorname{ar}(6)=\operatorname{ar}(3)+\operatorname{ar}(4)+\operatorname{ar}(5)
\end{aligned}
$

CL is the median

$
\begin{aligned}
& \operatorname{ar}(\triangle C A L)=\operatorname{ar}(\triangle C B L) \\
& \operatorname{ar}(1)+\operatorname{ar}(6)+\operatorname{ar}(5)=\operatorname{ar}(2)+\operatorname{ar}(3)+\operatorname{ar}(4)
\end{aligned}
$

BN is the median

$
\begin{aligned}
& \operatorname{ar}(\triangle B N A)=\operatorname{ar}(\triangle B N C) \\
& \operatorname{ar}(4)+\operatorname{ar}(5)+\operatorname{ar}(6)=\operatorname{ar}(1)+\operatorname{ar}(2)+\operatorname{ar}(3)
\end{aligned}
$

$
\begin{aligned}
& \text { (i) }- \text { (ii) } \\
& \operatorname{ar}(1)+\operatorname{ar}(2)+\operatorname{ar}(6)-\operatorname{ar}(1)-\operatorname{ar}(6)-\operatorname{ar}(5)=\operatorname{ar}(3)+\operatorname{ar}(4)+\operatorname{ar}(5)-\operatorname{ar}(2)- \\
& \operatorname{ar}(3)-\operatorname{ar}(4) \\
& \text { So, }{ }^{a r}(2)=\operatorname{ar}(5)
\end{aligned}
$

$
\begin{aligned}
& \text { (ii) }-(\mathrm{iii}) \\
& \operatorname{ar}(1)+\operatorname{ar}(6)+\operatorname{ar}(5)-\operatorname{ar}(4)-\operatorname{ar}(5)-\operatorname{ar}(6)=\operatorname{ar}(2)+\operatorname{ar}(3)+\operatorname{ar}(4)-\operatorname{ar}(1)- \\
& \operatorname{ar}(2)-\operatorname{ar}(3)
\end{aligned}
$

$
\operatorname{ar}(1)-\operatorname{ar}(4)=\operatorname{ar}(4)-\operatorname{ar}(1)
$

So, ${a r}(1)=\operatorname{ar}(4)$

Similarly, we can prove that area of $\operatorname{ar}(1)=\operatorname{ar}(2)=\operatorname{ar}(3)=\operatorname{ar}(4)=\operatorname{ar}(5)=\operatorname{ar}(6)$

Hence the triangle is divided into 6 triangles of equal area.
$\triangle A G B, \triangle A G C$ and $\triangle B G C$ consists of 2 triangles each

So they have equal area which is equal to twice the area of the smaller triangles. $\operatorname{ar}(\triangle A G B)=\operatorname{ar}(\triangle A G C)=\operatorname{ar}(\triangle B G C)=2$ (area of 1 small triangle)

Hence,

$
\operatorname{ar}(\triangle A G B)=\operatorname{ar}(\triangle A G C)=\operatorname{ar}(\triangle B G C)=\frac{1}{3} \operatorname{ar}(A B C)
$

Hence proved.

 

Posted by

infoexpert26

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