A Unique way to find the area of a triangle

Let ABC be a general triangle.let AD,BE,CF be the medians of triangle ABC Let G be the centroid...express the area of triangle ABC in terms of AD,BE,CF... HINT: extend AG to K such that AG=GK

New Incarnation to the problem...Let p(x) be a cubic polynomial over real such that AD,BE,CF are its roots. then prove that area of triangle ABC is 1/3\sqrt{(4((x)^2-2yz)-(S2)^2} where x =sigma 2 y=sigma one z=sigma 3 S2=sum of the squares of the roots.. Hint:theory of equations,previous problem

#Geometry #Area #Concurrency #Medians

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$