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.

  • $\begingroup$ Do you mean something like this? Ignore the extra lines. $M$ is the arbitrary point and the parallels are drawn in dotted lines. $\endgroup$
    – Guy
    Commented Apr 3, 2014 at 12:08
  • $\begingroup$ Yeah. But just to clear that the shaded regions in your diagram are not $S_1,S_2,S_3$ $\endgroup$
    – user140087
    Commented Apr 3, 2014 at 12:11
  • $\begingroup$ 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. $\endgroup$ Commented Apr 3, 2014 at 12:14
  • $\begingroup$ 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. $\endgroup$
    – Guy
    Commented Apr 3, 2014 at 12:14
  • 1
    $\begingroup$ 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... $\endgroup$
    – DonAntonio
    Commented Apr 3, 2014 at 12:17

2 Answers 2


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).

enter image description here

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

  • $\begingroup$ wow this is brilliant. $\endgroup$
    – Guy
    Commented Apr 4, 2014 at 9:20

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.


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