Chapter 15.3, Problem 6PSA

To determine

To Prove: Angle 1 is less than Angle 2.

Explanation of Solution

Given:

  $\begin{array}{l}\to CO\\ \to AB<CD\end{array}$

  

Concept Used:

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

Calculation:

  $\begin{array}{l}\to AB<CD\left(\text{Given}\right)\\ \text{⇒}OA=OB=OD=OC\left(\text{All are the radius originate from }O\right)\\ \text{⇒}\angle 1<\angle 2\left(\text{ SSS inequality theorem}\right)\end{array}$

Conclusion:

Thus,from the given condition and using SSS inequality theorem it is proved that $\angle 1<\angle 2$