Problem: Let $A$, $B$, $C$, $D$, and $E$ be points on a circle. For any three points, we draw the line going through the centroid of the triangle formed by these three points that is perpendicular to the line passing through the other two points. (For example, we draw the line going through the centroid of $\triangle BDE$ that is perpendicular to $\overline{AC}$.) In this way, we draw a total of $\binom{5}{3} = 10$ lines. Show that all 10 lines pass through the same point.
My work so far: I know the centroid of $\triangle ABC$ can be expressed as $\overrightarrow{G_1}=\frac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3}$. I believe the equation for the line that through from this point perpendicular to $\overrightarrow{DE}$ can be expressed as $\overrightarrow{l}=(\text{proj}_{\overrightarrow{DE}}\overrightarrow{G_1}-\overrightarrow{G_1})t+G_1$. Is this correct? If so, how can I use this to prove all 10 lines are congruent? Thank you in advance.
Note: By "$\text{proj}_{\overrightarrow{DE}}\overrightarrow{G_1}$" I mean the projection of $\overrightarrow{G_1}$ onto $\overrightarrow{DE}$.
Best Answer
claim: all ten lines go through the point $\frac{1}{3}(a+b+c+d+e).$
proof: i will use complex numbers instead of vectors. chooses a coordinate system so that the circle has unit radius and the point $D$ is at $1.$ let the complex numbers $a, b, c=e^{2i\gamma}, d = 1, e = e^{2i\epsilon}.$ the line through the center of the triangle $ABC$ and orthogonal to line $DE$ has the parametric form $$\mbox{ line1:} \frac{1}{3}(a+b+c) +\frac{s}{3}e^{i\epsilon} , s \mbox{ real}$$ in the same way the line through the center of the triangle $ABE$ orthogonal to $DC$ has the parametric form $$\mbox{ line2:} \frac{1}{3}(a+b+e) + \frac{t}{3}e^{i\gamma}, t \mbox{ real}$$
solving the two equation and their complex conjugates, we find that $$ s =e^{i\epsilon} + e^{-i\epsilon} \mbox{ and } t = e^{i\gamma} + e^{-i\gamma}$$ and the common points of intersection as claimed.