TITLE: Area of the given triangle
QUESTION [10 upvotes]: 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.

REPLY [3 votes]: You have three small triangles each of which is similar to the original triangle with the same orientation.  Their bases add up to the base of the original triangle (just slide two of them down to the bottom).

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 $S_0=\left(\sqrt{S_1}+\sqrt{S_2}+\sqrt{S_3}\right)^2$.