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Triangles between Same Parallels

If the median...

Question

If the medians of a $A B C$ intersect at $G$ show that $a r (Δ A G B) = ar (Δ A G C) = \frac{1}{3} ar (Δ A B C)$

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Solution

Given AM, BN and CL are the medians.

In $Δ A G B$ and $Δ A G C$

AG is the median

$∴$ Area $(Δ A G B) =$ area $(Δ A G C)$

Similarly, BG is the median

$∴$ Area $(Δ A G B) =$ area $(Δ B G C)$

So, area $(Δ A G B) =$ area $(Δ A G C) = a r e a Δ B G C$

Now, $Δ A G B + Δ A G C + Δ B G C =$ area $(Δ A B C)$

$\Rightarrow \frac{1}{3} Δ A G B + \frac{1}{3} Δ A G C + \frac{1}{3} Δ B G C =$ Area $(Δ A B C)$

So $a r (Δ A G B) =$ area $(Δ A G C) =$ area $(Δ B G C) = \frac{1}{3}$ Area $(Δ A B C)$

Hence proved.

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