Sum of areas of triangles Let  $ABC$ an  equilateral   triangle  and  $O$ a  point in  interior of $ABC$. Consider $M, N, P$  the  projections on  the  sides $AB, BC, AC$.  Then  the  sum of  areas of  triangles $ AOM, BON, COP$ is  the same like  the  sum  of  areas of  triangles $BOM, CON, AOP$. How  to  start? I  try  to  find  an  elementary  solution.
 A: 
Let $|AB|=|BC|=|CA|=a$,
$|AM|=m$,
$|BN|=n$,
$|CP|=p$,
$|OM|=u$,
$|ON|=v$,
$|OP|=w$.
\begin{align}
S_1&=
S_{AMO}+S_{BNO}+S_{CPO}
,\\
S_2&=
S_{AOP}+S_{BOM}+S_{CON}
,\\
S_1+S_2&=S_{ABC}
.
\end{align}
Given 
\begin{align}
A&=(0,0),\quad B=(a,0),\quad C=(\tfrac12a,\tfrac{\sqrt3}2a),\\
O&=(m,u),\quad M=(m,0),\\
\end{align}
using complex arithmetics,
it can be shown that 
\begin{align}
N&=(\tfrac34a-\tfrac{\sqrt3}4\,u+\tfrac14\,m
,\ 
\tfrac{\sqrt3}4\,a+\tfrac34\,u-\tfrac{\sqrt3}4\,m),\\
P&=(
\tfrac{\sqrt3}4\,u+\tfrac14\,m
,\ 
\tfrac34\,u+\tfrac{\sqrt3}4\,m),\\
S_{AMO} &= \tfrac12\,u\,m,\\
S_{BNO} &= \tfrac18\,(\sqrt3\,a-\sqrt3\,m-u)(a+\sqrt3\,u-m),\\
S_{CPO} &= \tfrac18\,(2\,a-m-\sqrt3\,u)(\sqrt3\,m-u),\\
S_1&=
S_{AMO}+S_{BNO}+S_{CPO}
=\tfrac{\sqrt3}8\,a^2=\tfrac12 S_{ABC}
=S_2
.
\end{align}
A: COMMENT.-You can demonstrate your problem if you are willing to "perspire" enough. The calculations are direct but tedious. I have tried, to help you, to verify the property with a numerical example, less laborious than with algebraic variables literal and would seem a counterexample because I have not found the equality of the sums. It could be that because the distances ON, OM and OP I have calculated with only a few decimal places. Anyway, here the numerical example which you could examine in more detail (if you want of course).
The equilateral triangle with vertices $A(0,0),B(4,0),C(2,2\sqrt3)$, the interior point $0=(3,1)$.
You get $$BN=1.366008053\space \space \space CP=1.634013948\\ON=0.366030053\space \space \space OP=2.098068159$$ We take the double of areas
$$\triangle AOM=3\\\triangle BON=0.500000000038016809\\\triangle COP=3.428272635660681732$$
Their sum is $$S_1=6.928272635698698541$$
$$\triangle BOM=1\\\triangle CON=0.964120211961983191\\\triangle AOP=4.964000000339318268$$
 whose sum is equal to $$S_2=6.928120212301301459$$
