$\vec{PP_1}+\vec{PP_2}+\vec{PP_3}=\frac {3}{2}\vec {PO} $. Let $ABC $ an equilateral  triangle and $P\in int (ABC) $. 
If $O $ is the center of gravity of  the triangle and $P_1, P_2, P_3$ are the projections of $P $ on the sides then 
$\vec{PP_1}+\vec{PP_2}+\vec{PP_3}=\frac {3}{2}\vec {PO} $.
 A: Let 
$A_1C_2||AC $, $A_2B_1||AB $ and $B_2C_1||BC $ trough $P $
where $A_1,B_2\in [AB] $, $A_2, C_1\in [AC ]$ and $B_1, C_2\in [BC] $.
Then $PA_1AA_2$,  $PC_1CC_2$,  $PB_1BB_2$ are parallelograms and $A_1PB_2$, $A_2PC_1$, $B_1PC_2$ are equilateral triangles.
Now,
$\vec {PP_1}=\frac {\vec{PB_1}+\vec{PC_2}}{2}$
$\vec {PP_2}=\frac {\vec{PA_2}+\vec{PC_1}}{2}$
$\vec {PP_3}=\frac {\vec{PA_1}+\vec{PB_2}}{2}$.
But 
$\vec {PA_1}+\vec {PA_2}=\vec {PA} $
$\vec {PB_1}+\vec {PB_2}=\vec {PB} $
$\vec {PC_1}+\vec {PC_2}=\vec {PC} $.
So,
$\vec {PP_1}+\vec {PP_2}+\vec {PP_3}=\frac {\vec {PA}+\vec {PB}+\vec {PC}}{2}=\frac {3\vec {PO}}{2} $, q.e.d.
A: Given the vertices $A,B,C$ then $O=\frac{A+B+C}3$ which is a weighted sum with equal weights. Now $P=xA+yB+zC$ for some weights $x,y,z$ such that $x+y+z=1.$ Suppose $P_1$ is the projection of $P$ on side $AB$ with midpoint $M=\frac{A+B}2$. Then $P_1-P= M-(xM+yM+zC)$ and similarly for the other two projections. Calculation shows that $\frac{P_1+P_2+P_3}3-P=\frac{O-P}2.$
Note that this proof uses affine geometry which works for any triangle but we need to interpret projection onto a side as parallel projection along the same direction as the median onto that side.
A: 
Without loss of generality let $\overline{AB} = 1$. Let $\overline{AM} = y$ and $\overline{AN} = x$, where $\overline{PN}$ is parallel to $\overline{AB}$ and $\overline{PM}$ is parallel to $\overline{AC}$. Assume $x\geq y$. Note that to have $P$ lie in the interior, you need $x+y \leq 1$.
It is readily checked that $\overline{BP'} = \frac{1}{2}(1-y+x)$ and $\overline{CP'} = \frac{1}{2}(1+y-x)$. So $\overline{O'P'} = \frac{1}{2}(x-y)$. 
We can also see that $\overline{O'P_1'} = \frac{1}{2}(y + \frac 12 x)$ and $\overline{O'P_2'} = \frac{1}{2}(x+\frac12 y)$, so $\overline{P'P_1'} = \frac{3x}{4}$ and $\overline{P'P_2'} = \frac{3y}{4}$.
Hence $\overline{P'P_1'} - \overline{P'P_2'} = \frac32 \overline{P'O'}$. You need to check why this implies $\vec{PP_1} + \vec{PP_2} + \vec{PP_3} = \frac32 \vec{PO}$.
