Let one of the vertices be $O(0,0)$ and let the other vertices $A_1, \dots, A_{n-1}$ lie on a circle with radius $1$ and centre $O$ in such a way that $A_1(1,0)$, $A_{n-1}(0,1)$ and $A_2, \dots, A_{n-2}$ are all on the shorter arc $A_1A_{n-1}$ (Fig. 16 depicts the case $n=6$).

The vertex farthest from line $OA_1$ is $A_{n-1}$, the vertex farthest from line $OA_{n-1}$ is $A_1$. The vertex farthest from any other line determined by a side of the polygon is $O$ because the line passing through $O$ parallel to such a side lies in II and IV quarters while the other vertices of the polygon lie above it in I quarter. Thus the polygon is correct. The altitudes drawn to sides $OA_1$ and $OA_{n-1}$ are $OA_{n-1}$ and $OA_1$, respectively, they meet at point $O$. As $O$ is the vertex farthest from any other side, all other altitudes meet in $O$, too.

Fig. 16