Chapter 15.2, Problem 6PSA

To determine

To Prove: The angle is greater than the other in the given figure.

Explanation of Solution

Given:

  $\begin{array}{l}\angle 1\text{ is the exterior angle of }\Delta \text{ACD}\text{.}\\ \angle \text{2 is the exterior angle of }\Delta \text{ABC}\text{.}\end{array}$

  

Formula Used:

Exterior Angle of a triangle is always greater than its remote interior angles.

Calculation:

  $\begin{array}{l}\angle 1>\angle 2\left(\text{Exterior angle of a triangle > each of its remote interior angles}\right)\\ \angle 2>\angle 3\left(\text{Exterior angle of a triangle > each of its remote interior angl}es\right)\\ hence\\ \angle 1>\angle 3\end{array}$

Conclusion:

Hence, Proved.