Chapter 15.3, Problem 12PSB

To determine

To Prove: The side BC is greater than the side AD in the given figure.

Explanation of Solution

Given:

  $\begin{array}{l}\to \angle 1=\angle 3\\ \to \overline{BA}\parallel \overline{CD}\\ \to AC>AD\end{array}$

  

Concept Used:

The Hinge Theorem: It states that if two sides of two triangles are congruent and angle of one is greater than the angle of second then the third side of the triangle will be greater than the another.

Calculation:

  $\begin{array}{l}\text{⇒}\overline{BA}\parallel \overline{CD}\left(\text{Given}\right)\\ \text{⇒}\angle 2\cong \angle 3\left(\begin{array}{l}\text{Two parallel line cut by transversal,each pair of alternate}\\ \text{interior angle are congruent}\text{.}\end{array}\right)\\ \text{⇒}\angle 1<\angle 3\left(\text{Given}\right)\\ \text{⇒}\angle 1<\angle 2\left(\text{Because }\angle 2\cong \angle 3\right)\\ \text{⇒}BC>AC\left(\begin{array}{l}\text{Two side of a triangle are not congruent,then the angles opposite}\\ \text{them are not congruent and the larger angle is opposite the longer side}\end{array}\right)\\ \text{⇒}AC>AD\left(\text{Given}\right)\\ \text{⇒}BC>AD\left(\text{Transitive property of inequality and }BC>AC\text{ and }AC>AD\right)\end{array}$

Conclusion:

Thus, it is proved that $BC>AD$