Your solution takes two lines, calculates their intersection, and confirms that it is as you claimed. Here is a proof which uses position vectors to show that each line goes through the specified point $$\mathbf{p}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}+\mathbf{e}}{3}.$$ I note at the end that this proof works as well if their are more points (with a slight change) and also in $3$ or more dimensions. The proof certainly works, but I still am not satisfied that it is as clear as it could be.
The claim is that, given 5 points $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d},\mathbf{e}$ on a circle $\mathbf{x} \cdot \mathbf{x}=r^2$ (i.e. centered at the origin), the line through the point $\mathbf{p}$ and the centroid $$\mathbf{q}=\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}3$$ of the triangle determined by the first three is perpendicular to the line through the points $\mathbf{d}$ and $\mathbf{e}.$ The first line, $\overline{\mathbf{pq}},$ is in the direction of the vector $\mathbf{d}+\mathbf{e}$ and the second line, $\overline{\mathbf{de}},$ is in the direction of the vector $\mathbf{d}-\mathbf{e}.$ The dot product of these two direction vectors is $\mathbf{d}\cdot \mathbf{d}-\mathbf{e}\cdot \mathbf{e}=r^2-r^2=0.$ Hence they are, indeed, perpendicular.
Perhaps something even more direct is possible.
Generalization:
Given $n+2$ points $\mathbf{a_1},\mathbf{a_2},\cdots,\mathbf{a_{n+2}},$all equally distant from the origin, consider all $\binom{n+2}{2}$ lines determined by choosing $n$ of the points and taking the line through their centroid which intersects and is perpendicular to the line determined by the other $2.$ These lines all pass through the point $$\frac{\mathbf{a_1}+\mathbf{a_2}+\cdots+\mathbf{a_{n+2}}}{n}$$.