Problem with Equilateral Triangle Randomly choose a point $P$ in the interior of an equilateral triangle $ABC$, which has a side of length $a$. Let $D,E,F$  be the feet of the perpendiculars from $P$ to $BC,CA,AB$ respectively. Show that $(PDC)+(PFB)+(PEA)=\frac{a^2\sqrt3}{8}$.  
Here are my thoughts so far:
We want to show $(PDC)+(PFB)+(PEA)=\frac{ABC}{2}$, therefore it suffices to show that $(PDC)+(PFB)+(PEA)=(PDB)+(PFA)+(PEC)$  
I believe that this can be shown using the following two relations:
1)$PD+PE+PF=$height of $ABC$ (Viviani's Theorem)
2)$BD+CE+AF=DC+EA+FB=3a/2$
Here is a proof of 2): By the Pythagorean theorem we get
$(a-BD)^2+PD^2=CE^2+PE^2$
$(a-CE)^2+PE^2=AF^2+PF^2$
$(a-AF)^2+PF^2=BD^2+PD^2$
Addition of the previous three equation yields the result.
The problem now is if and how 1) and 2) can lead to  a solution.
 A: Unfortunately, getting a proof from the two equations reported in the OP seems difficult, since the heights of the three triangles are all different. 
The proof can be achieved using an alternative method. Orient an equilateral triangle with side $s$ in a way that $A$ is the vertex at the top, and the base $BC$ is at the bottom. As described in the OP, let us choose an internal point $P$, and let us call $D,E,F$ the feet of the perpendiculars from $P$ to $BC$, $AC$, and $AB$, respectively. Consider the height $AD$ normal to $BC$. It is trivial to show that, if the point $P$ is randomly chosen on $AN$, then the sum of the areas of the triangles $PDC$, $PFB$, and $PEA$ is equal to half of the total area of the triangle $ABC$, since this is divided in three pairs of symmetric triangles.
Keeping the point $P$ on $AD$, let us set $PD=d$, $BF=CE=b$, and $PF=PE=h$. The triangle $PDC$ has base equal to $\frac{s}{2}$ and heigth equal to $d$. The triangle $PFB$ has base $b$ and height $h$. The triangle $PEA$ has base $s-b$ and height $h$. We can note that $h=\frac{(s \frac{\sqrt{3}}{2}-d)}{2}$ (in fact, looking at the right triangle $PAF$, we have $PA=s \frac{\sqrt{3}}{2}-d$ and angle $PAF=30^o$). By similar considerations, it can also be shown that $b=  \frac{2 \sqrt{3}}{3}d + \frac{(s \frac{\sqrt{3}}{2}-d)\sqrt{3}}{6}$.  
Now let us move the point $P$ laterally in a direction normal to $AD$ (e.g., leftward) by a distance $x$. The area of the triangle $PDC$ increases by $\frac{dx}{2}$. The area of the triangle $PFB$ changes by $\frac{1}{2}[(b-x/2)(h-x \sqrt{3}/2)-bh]$ (note that the change is negative). The area of the triangle $PEA$ changes by $\frac{1}{2}[(s-b-x/2)(h+x \sqrt{3}/2)-(s-b)h]$ (here the change can be positive or negative, because the base is reduced and the height increases). Summing all changes we have that the overall change $C$ is
$$C=\frac{dx}{2}+\frac{1}{2}[(b-x/2)(h-x \sqrt{3}/2)-bh]+\frac{1}{2}[(s-b-x/2)(h+x \sqrt{3}/2)-(s-b)h]= \\ \frac{dx}{2} +\frac{sx \sqrt{3}}{4}-\frac{bx\sqrt{3}}{2}-\frac{hx}{2}$$
Substituting $b$ and $h$ with the equations reported above, we get
$$C=\frac{dx}{2} +\frac{sx \sqrt{3}}{4}- dx -\frac{sx \sqrt{3}}{8}+\frac{dx}{4}-\frac{sx \sqrt{3}}{8}+\frac{dx}{4}=0$$
which shows that, moving the point $P$ leftward by  any distance $x$, the overall change in the sum of the areas of the triangles $PDC$, $PFB$, and $PEA$ is zero. Because of the symmetry of the problem, the same conclusion can be drawn if we consider a rightward movement.
Since any point inside the triangle $ABC$ can be reached starting from a given point of the height $AD$ (defined by a value of $d$) and moving laterally by a distance $x$, we can conclude that the sum of the areas of the triangles $PDC$, $PFB$, and $PEA$ is constant whatever the point chosen, and then is equal to $s^2 \frac{\sqrt{3}}{8}$.
