In a quadrilateral ABCD, the diagonals AC and BD intersect at O.
Using the triangle inequality on the four triangles formed, AC + BD is less than which sum?

In triangles AOB, BOC, COD, DOA: OA+OB>AB, OB+OC>BC, OC+OD>CD, OD+OA>DA. Adding gives 2(OA+OB+OC+OD) > perimeter, i.e. AC+BD > $\frac{\text{perimeter}}{2}$; and separately AC+BD < AB+BC+CD+DA from the triangle inequalities on triangles ABC and ACD etc. So AC + BD < AB + BC + CD + DA.