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.
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 $P$.
Best Answer
WLOG, say the center of the circle ($O$) is at the origin. Vertices of the pentagon $ABCDE$ are represented by position vectors $\overline{a}, \overline{b}, \overline{c}, \overline{d}$ and $\overline{e}$.
Centroid of $\triangle ABC, \, \overline {g} = \frac{\overline{a} + \overline{b} + \overline{c}}{3}$
Line $DE = \overline{d} - \overline{e}$
As points $A, B, C, D, E$ are concyclic with center at $O$
$|\overline{a}|^2 = |\overline{b}|^2 = |\overline{c}|^2 = |\overline{d}|^2 = |\overline{e}|^2$ ...(i)
If a point $P$ with position vector $\overline{p} \,$ is on the perpendicular line from the centroid of $\triangle ABC$ to the line $DE$,
$(\overline{p}-\overline{g}) \cdot (\overline{d} - \overline{e}) = 0$
Based on (i) one of the ways for the dot product to be zero is
$(\overline{p}-\overline{g}) = n_1 (\overline{d}+\overline{e}) \,$ (you can easily show why $\overline{p} = \overline{g}$ will not give you the concurrent point by symmetry)
$\overline{p}-\overline{g} = \overline{p}-\frac{\overline{a} + \overline{b} + \overline{c}}{3} = n_1 (\overline{d}+\overline{e})$ ...(ii)
Similarly,
$\overline{p}-\frac{\overline{b} + \overline{c} + \overline{d}}{3} = n_2 (\overline{e}+\overline{a})$ ...(iii)
From (ii)-(iii), you get one solution when $n_1 = n_2 = \frac{1}{3}$ and
$\overline {p} = \frac{\overline{a} + \overline{b} + \overline{c} + \overline{d} + \overline{e}}{3}$
Now we need to prove this point is the point of concurrency for other $3$ lines too. So we take the lines from centroids of $\triangle CDE, \triangle DEA, \triangle EAB$ through point $\overline {p}$ and show each of them is perpendicular to the line segment made by other two vertices.
$(\overline{p}- \frac{\overline{c} + \overline{d} + \overline{e}}{3}) \cdot (\overline{a} - \overline{b}) = 0$
$(\overline{p}- \frac{\overline{d} + \overline{e} + \overline{a}}{3}) \cdot (\overline{b} - \overline{c}) = 0$
$(\overline{p}- \frac{\overline{e} + \overline{a} + \overline{b}}{3}) \cdot (\overline{c} - \overline{d}) = 0$
which is easy to show given (i).