Chapter 15.2, Problem 16PSB

To determine

To Prove: The sum of three sides of quadrilateral is greater than the third side.

Explanation of Solution

Given:

  $AB$ $\text{CD}$ is a quadrilateral.

  $AB+BC+CD>AD$

Formula Used:

  $\begin{array}{l}\text{Transitive Property of Triangle}\\ If a < b and b < c ,then a < c.\\ If a \le b and b \le c ,then a \le c.\\ If a > b and b > c ,then a > c.\\ If a \ge b and b \ge c ,then a \ge c.\end{array}$

Calculation:

  $\begin{array}{l}Drawanauxiliarysegment\text{}\text{ BD}\\ \text{in }\Delta \text{BCD}\\ BC+CD>BD\\ AB+BC+CD>AB+BD\left(\text{additional property of inequality}\right)\\ in\Delta ABD\\ AB+BD>AD\left(\text{triangle inequality postulate}\right)\\ by\text{ equalising}\\ AB+BC+CD>AD\\ \text{hence proved}\text{.}\end{array}$

Conclusion:

Hence Proved.