It is possible to give a syhtnetic proof using the following result:
The perpendiculars dropped from the midpoints of a cyclic quadrilateral to the opposite sides are concurrent. Furthermore, their point of intersection is the symmetric of the center of the circumscribed circle with respect to the centroid of the quadrilateral.
Proof: Let $M,N$ be the midpoints of $AB,CD$ and denote $X,Y$ the projections of $M,N$ on $CD,AB$, respectively. Denote by $Z$ the intersection of $MX,NY$. Denote by $O$ the center of the circle. The perpendicularity condition shows that $OMZN$ is a parallelogram, and thus, $OZ,MN$ have the same midpoint. But the midpoint of $MN$ is the centroid of $ABCD$. It is straightforward to see that the other two perpendiculars also pass through $Z$. $\square$
Returning to the initial configuration of the problem, choose a point $P$ among the $5$ points and consider a homothety $h$ of ratio $3/2$ of center $P$. Denote the rest of the points $A,B,C,D$. The line through the centroid of $PAB$ perpendicular to $CD$ is mapped to the line which passes through the midpoint of $AB$ and is perpendicular on $CD$. Thus the four lines corresponding to triangles which contain $P$ are mapped by $h$ into four concurrent lines like in the result proven above. Because the homothety preserves concurrency, the four lines associated to $P$ are also concurrent. For a point $P$ we denote $P_a,P_b,P_c,P_d$ these concurrent lines.
It is not hard to see that if $P,Q$ are different then $\{P_a,P_b,P_c,P_d\} \cap \{Q_a,Q_b,Q_c,Q_d\}$ has at least two elements, thus the concurrency point is the same for every $4$-family of lines associated to a point on the circle. This shows that the ten lines are concurrent.
Here is a Geogebra configuration.
If you want, you can find the precise location of the intersection point, but for this, the fastest way is using complex numbers, and this has already been done.