Question

If O is a point within a quadrilateral ABCD, show that OA + OB + OC + OD > AC + BD.

Answer 1

Let ABCD be a quadrilateral whose diagonals are AC and BD and O is any point within the quadrilateral.
Join O with A, B, C, and D.
We know that the sum of any two sides of a triangle is greater than the third side.
So, in ∆AOC, OA + OC > AC

Also, in ∆ BOD, OB + OD > BD

Adding these inequalities, we get:

(OA + OC) + (OB + OD) > (AC + BD)
⇒ OA + OB + OC + OD > AC + BD