Dave cracket this toughnut for us using position vectors! Well done! You might like to consider whether the result holds in three dimensions when the five given points are not coplanar.

Consider the pentagon:

We can obtain 5 simultaneous equations:
2a =A+B
2b =B+C
2c =C+D
2d =D+E
2e =E+A

Solving these equations gives:
A=a -b +c -d +e
Similar formulas for B, C, D and E are also obtained.

Note this formula can also be written as:
A=c +(e -d )+(a -b )

So if we start at C, then construct a line of the length of the line from e to d parallel to it, starting from c , then a line starting from the previous point the same length and direction as the line from b to a , then this final point will be A.

But how do you construct a parallel line through a point z, parallel to the line through x, y and the same length?

Draw a line connecting x to z and beyond, then draw a circle with radius less than the distance to z , centred on x . Draw a circle with the same radius centred on z. Now set the compass to the length from a to b, and raw a circle about c with this radius. Now if you draw a line through where this circle intersects our previous circle, you will have a line through z parallel to the line through x and y . Finally, to get the length correct, set the compass to the length from x to y , and draw a circle of this radius about z . The intersection with the line gives the endpoint.

In the case of a general odd-sided shape, the formula will just be:

X1=x1-x2+x3-...+xn
and similar formulas will apply for all the other points. by the same factorisation as above, a combination of lines can again be used to find each point.

If we have an odd number of sides, when we try and solve the equations we find:

x1-x2+x3-...+xn-1-xn=0
If this is true, then we can solve the problem form even-sided shapes, although there will be multiple possible solutions this time - we can choose one of the verticies to be whatever we like!