Chapter 6.4, Problem 18WE

To determine

To Prove : The expression, $AB+BC+CD+DA>2\left(AC\right)$ for the given figure.

Explanation of Solution

Given information:

  

The sum of the length of any two sides of a triangle is greater than the length of the third side

. $\begin{array}{l}AB+BC>AC\hfill \\ AB+AC>BC\hfill \\ AC+BC>AB\hfill \end{array}$

  

First draw the diagram to scale as below:

  

Use of the above theorem,

  $\begin{array}{l}AB+BC>AC\mathrm{.......}\left(1\right)\hfill \\ CD+DA>AC\mathrm{........}\left(2\right)\hfill \end{array}$

Add equations (1) and (2) to get

  $AB+BC+CD+DA>2\left(AC\right)$

Hence, $AB+BC+CD+DA>2\left(AC\right)$ . (Proved).