Vector Applications in Euclidean Geometry Pentagon $ABCDE$ is inscribed in a circle. For any edge of $ABCDE$, we can draw the line perpendicular to that edge that contains the centroid of the remaining three vertices. Show that these 5 lines are concurrent.
We just finished a unit on vectors in my Pre-Calculus class but I have no idea on how to do this final problem. Any help would be greatly appreciated!

 A: WOLOG, assume the pentagon is inscribed in a circle of radius $R$ centered at origin $O$.
Let $p_1, p_2, p_3, p_4, p_5$ be the vectors correspond to vertices $A,B,C,D,E$
respectively. We have
$$|p_1|^2 = |p_2|^2 = |p_3|^2 = |p_4|^2 = |p_5|^2 = R^2$$
Let $G$ be the vertex centroid of the pentagon and $P$ be a point at
the five-third mark on the ray $OG$. If $p$ is the corresponding vector, we have
$$p = \frac53 \left(\frac{p_1 + p_2 + p_3 + p_4 + p_5}{5}\right) = \frac{p_1 + p_2 + p_3 + p_4 + p_5}{3}$$ 
Let $[5] = \{ 1, 2, 3, 4, 5 \}$. For any distinct $i, j \in [5]$, let $\ell, m, n$ be the rest of indices from $[5]$.
i.e. $[5]$ is a disjoint union of $\{ i, j \}$ and $\{ \ell, m, n \}$.
We have
$$(p_i - p_j) \cdot \left( p - \frac{p_\ell + p_m + p_n}{3}\right)
= \frac{(p_i - p_j) \cdot (p_i + p_j)}{3} = \frac{|p_i|^2 - |p_j|^2}{3} = 0$$
This is the equation for point $p$ lying on a line passing through
$\frac{p_\ell + p_m + p_n}{3}$ (the centroid of triangle with vertices $p_\ell$, $p_m$, $p_n$ ) perpendicular to line $p_i p_j$.
Substitute $p_i,p_j$ by $p_1,p_2$, we find $P$ lie on a line
perpendicular to $AB$ which passes through the centroid of triangle $CDE$.
Same thing happens to other sides of the pentagon. 
As a result, the $5$ lines mentioned in question are concurrent and meet at this specific point $P$.
