Draw quadrilateral ABCD.
If the convex hull is a triangle, then the fourth point lies inside its circumcircle.
So suppose ABCD is convex. One pair of opposite angles must have sum greater than 180° (otherwise the points would lie on a circle; if we take four points on a circle, opposing angles always sum 180°).
Suppose they are A and C. Then we claim that C lies inside the circle through A, B, D. Take a point C' on the ray DC on the far side of C from D such that:
m
Then, m
So, C' lies on the circle.
Hence C, which lies on the segment C'D, lies inside the circle. |