Geometry and Trigonometry

Abstract

Problem 2.5 Let ABCD be a convex quadrilateral. Prove that

$$\max(AB+CD,AD+BC)<AC+BD<AB+BC+CD+DA.$$

Solution Let O be the point of intersection of the diagonals AC and BD. We have AO+OB>AB and CO+OD>CD; thus AC+BD>AB+CD. Similarly, AO+OD>AD and BO+OC>BC; thus AC+BD>AD+BC. It follows that

$$\max(AB+CD,AD+BC)<AC+BD.$$

For the second inequality, note that AC<AB+BC and AC<AD+DC; hence

$$AC<\frac{1}{2}(AB+BC+CD+DA).$$

Analogously,

$$BD<\frac{1}{2}(AB+BC+CD+DA),$$

and the result follows by adding these inequalities.