Question

In a quadrilateral ABCD, show that (AB + BC + CD + DA) > (AC + BD).

Answer 1

Given: Quadrilateral ABCD

To prove: (AB + BC + CD + DA) > (AC + BD)

Proof:

$$\begin{gathered} \mathrm{In}∆ABC, \\ AB+BC>AC...\left(\mathrm{i}\right) \\ \mathrm{In}∆CAD, \\ CD+AD>AC...\left(\mathrm{ii}\right) \\ \mathrm{In}∆BAD, \\ AB+AD>BD...\left(\mathrm{iii}\right) \\ \mathrm{In}∆BCD, \\ BC+CD>BD...\left(\mathrm{iv}\right) \end{gathered}$$

Adding (i), (ii), (iii) and (iv), we get

2(AB + BC + CD + DA) < 2( AC + BD)

Hence, (AB + BC + CD + DA) < (AC + BD).