# Area of the given triangle

Through an arbitrary point lying inside a triangle, three straight lines parallel to its sides are drawn. These lines divide the triangle into six parts, three of which are triangles. If the areas of these triangles are $S_1,S_2,S_3$, then prove that the area of the given triangle equals $(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3})^2$.
Seriously, I have no idea how to approach this problem. It looks kind of complicated. Help and thanks in advance.

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Do you mean something like this? Ignore the extra lines. $M$ is the arbitrary point and the parallels are drawn in dotted lines. – Sabyasachi Apr 3 '14 at 12:08
Yeah. But just to clear that the shaded regions in your diagram are not $S_1,S_2,S_3$ – user140087 Apr 3 '14 at 12:11
I guess the triangles are $A'oM$, $B'oM$ and $C'oM$ with $o$ the unnamed points on the same sides as the first points. – gammatester Apr 3 '14 at 12:14
Yeah I know. This question just reminded of a question I had seen, thought it was the same, when I checked I saw that it is not the same. Good question though. – Sabyasachi Apr 3 '14 at 12:14
Well, in the meantime one point to consider and take into account: all those three triangles are similar to each other and to the big triangle as well... – DonAntonio Apr 3 '14 at 12:17

Call the base of the original triangle $b_0$ and its area $S_0$. Then $b_0=b_1+b_2+b_3$ and $S_i=kb_i^2$ with the same constant $k$, meaning that $\sqrt\frac{S_0}{k}= \sqrt\frac{S_1}{k}+\sqrt\frac{S_2}{k}+\sqrt\frac{S_3}{k}$ or in other words $S_0=\left(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3}\right)^2$.
There is a simple way to do this if you are prepared to use properties of affine mappings. First consider the case of the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$. Then one can compute the vertices of the three triangles very easily: they are $(x,y)$, $(1-y,y)$ and $(x,1-x)$, resp. $(x,y)$, $(x,0)$ and $(x+y,0)$, resp. $(x,y)$, $(0,y)$ and $(0,x+y)$. The triangle areas are then $x^2$, $y^2$ and $(1-x-y)^2$ up to a factor of $\frac 12$ which gives the required result. Now you use the fact that ANY triangle is affinely equivalent to the above one and that an affine mapping preserves parallelism and areas (up to a factor) and this allows you to reduce the general case to this one.
If you feel uncomfortable with this argument, you can choose coordinate system so that $A$ is $(0,0)$, $B$ is $(1,0)$ and $C$ is $(p,q)$. A similar, but slightly more complicated, computation then leads to the result.