
Geometry  Area
Hi,
I've been working on this problem all night and I don't know what I'm doing wrong! I keep on going in circles, I think I'm either missing something or making it to hard. I'm assuming P can be anywhere...
Problem:
P is a point inside an equilateral triangle ABC. Perpendiculars PD, PE AD PF are dropped from P onto BC, CA and AB respectively Prove that [PAF]+[PBD]+[PCE]=[PAE]+[PCD]+[PBF]
Thanks!

This argument is long and messy, but it works. I haven't been able to find a shorter proof.
Suppose for convenience that the sides of the triangle are 1 unit in length. The first thing to notice is that the sum of the distances from P to the three sides is equal to √3/2.
. . . . . . . . . . . . . . . . . . . .http://www.qbyte.org/puzzles/p117s1.gif
In this picture (gacked from here, where there is a full explanation), the area of the whole triangle is (1/2)base×height, and the height is √3/2. But this is equal to the sum of the three coloured triangles, whose bases are all equal to 1 and whose heights are $\displaystyle h_1,\;h_2$ and $\displaystyle h_3$. So $\displaystyle h_1 + h_2 + h_3 = \sqrt3/2$.
Notice that the area of the triangle ABC is (1/2)base×height = √3/4.
What we want to show here is that in the next picture (cunningly Photoshopped by me) the sum of the areas of the darkercoloured triangles is the same as the sum of the areas of the lightercoloured triangles.
. . . . . . . . . . . . . . . . . . . .http://www.thelances.co.uk/imagesC/triangle.png
Call the bases of the darkercoloured triangles $\displaystyle b_1,\;b_2$ and $\displaystyle b_3$. Then their total area is $\displaystyle {\textstyle\frac12}(b_1h_1+b_2h_2+b_3h_3)$.
The first main step is to find expressions for $\displaystyle b_1,\;b_2$ and $\displaystyle b_3$ in terms of $\displaystyle h_1,\;h_2$ and $\displaystyle h_3$.
For $\displaystyle b_2$, the vertical height of the point E above the horizontal line BC is $\displaystyle b_2\cos(\pi/3) = {\textstyle\frac{\sqrt3}2}b_2$. But it is also equal to $\displaystyle h_1+h_2\cos(\pi/6) = h_1+{\textstyle\frac12}h_2$. That leads to the equation $\displaystyle b_2 = {\textstyle\frac1{\sqrt3}}(2h_1+h_2)$.
Similarly, the vertical height of F above the line BC is equal to √3/2 (the height of A) minus the vertical drop from A to F. It is also equal to the drop from F to P plus the drop from P to D. This leads to the equation $\displaystyle b_3 = 1  {\textstyle\frac1{\sqrt3}}(2h_1+h_2)$.
For $\displaystyle b_1$, the distance DC is equal to $\displaystyle 1b_1$ and also to $\displaystyle {\textstyle\frac{\sqrt3}2}h_2 + {\textstyle\frac12}b_2 ={\textstyle\frac{\sqrt3}2}h_2 + {\textstyle\frac1{2\sqrt3}}(2h_1+h_2)$. This gives the equation $\displaystyle b_1 = 1  {\textstyle\frac1{\sqrt3}}(h_1+2h_2)$.
Let A denote the sum of the areas of the darkercoloured triangles. Then
. . . . . . . . . . . . . . .$\displaystyle \begin{array}{rcl}2A&=& h_1b_1 +h_2b_2 + h_3b_3 \\
&=& h_1(1  {\textstyle\frac1{\sqrt3}}(h_1+2h_2)) + h_2{\textstyle\frac1{\sqrt3}}(2h_1+h_2) + h_3(1  {\textstyle\frac1{\sqrt3}}(2h_1+h_2)) \\
&=& h_1 + h_3 + {\textstyle\frac1{\sqrt3}}\bigl[h_1^2 + h_2^2  2h_1h_3  h_3^2\bigr] \\
&=& h_1 + h_3 + {\textstyle\frac1{\sqrt3}}\bigl[h_2^2  (h_1+h_3)^2\bigr] .\end{array}$
Now use the fact that $\displaystyle h_1 + h_2 + h_3 = \sqrt3/2$ to write this as
. . . . . . . . . . . . . . . $\displaystyle \begin{array}{rcl}2A&=& h_1 + h_3 + {\textstyle\frac1{\sqrt3}}\bigl[h_2^2  ({\textstyle\frac{\sqrt3}2}  h_2)^2\bigr] \\
&=& h_1 + h_3 + {\textstyle\frac1{\sqrt3}}\bigl[{\textstyle\frac34} + \sqrt3h_2\bigr] \\
&=& h_1 + h_2 + h_3  {\textstyle\frac{\sqrt3}4} \\
&=& \textstyle\frac{\sqrt3}2  \textstyle\frac{\sqrt3}4 = \textstyle\frac{\sqrt3}4, \end{array}$
which is the area of the triangle ABC. Thus the sum of the areas of the darkercoloured triangles is half the area of the whole triangle, which is what we wanted to show.