Question

Show that in a quadrilateral ABCD AB + BC + CD + DA > AC + BD.
Figure

Answer 1

ABCD is a quadrilateral.
AC and BD are diagonals.

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{two}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}\mathrm{greater}\mathrm{than}\mathrm{the}\mathrm{third}\mathrm{side}. \\ \mathrm{In}△ACB,\mathrm{we}\mathrm{have}: \\ AB+BC>AC...\left(1\right) \\ \mathrm{In}△BDC,\mathrm{we}\mathrm{have}: \\ BC+CD>BD...\left(2\right) \\ \mathrm{In}△ACD,\mathrm{we}\mathrm{have}: \\ AD+DC>AC...\left(3\right) \\ \mathrm{In}△BAD,\mathrm{we}\mathrm{have}: \\ AB+AD>BD...\left(4\right) \\ \mathrm{Adding}\left(1\right),\left(2\right),\left(3\right)\&\left(4\right),\mathrm{we}\mathrm{get}: \\ AB+BC+BC+CD+AD+DC+AB+AD>AC+BD+AC+BD \\ \Rightarrow 2AB+2BC+2CD+2AD>2AC+2BD \\ \Rightarrow AB+BC+CD+AD>AC+BD \end{gathered}$$