Triangles and Medians

[Labib]

Find the ratio of the area of a given triangle to that of a triangle whose sides have the same lengths as the medians of the original triangle.

[sourav das]

First let's use a trick
Trick:

Just try it on a equilateral triangle and use the theorem that the ratio of area of two similar triangle is equal to the ratio of area of squares of similar sides.( This trick is not my own way , i get it from a bhaiia)

Now the direct proof
Direct solution:

Draw a figure where triangle $ABC$ has $AD, BE, CF$ medians meet at $G$ . Join $F$ and the mid point of $BG$(Suppose P).
Let's assume that the area of triangle made by the medians is $X$.
Now,
\[PF = \frac{1}{2}AG = \frac{1}{2}*\frac{2}{3}AD = \frac{1}{3}AD\Rightarrow \frac{PF}{AD} = \frac{1}{3} Similarly, \frac{FG}{CF} = \frac{1}{3} and \frac{GP}{BE} = \frac{1}{3}\]
So,\[\frac{X}{(PGF)} = 9\]
Because the theorem that the ratio of area of two similar triangle is equal to the ratio of area of squares of similar sides.
But,\[\frac{(ABC)}{(PGF)} = 12\]
And that's why: \[\frac{(ABC)}{X} = \frac{4}{3}\]

It is my direct proof

[Labib]

LAAL VAI, First of all, your trick is a well known one. You can find it in

class 10 higher maths book(theorem 3.10)

.

But, you know what? This is one hell of a proof!!! A fabulous one!!!