Chapter 15.3, Problem 11PSB

To determine

To Prove: In the given figure angle ABD is greater than Angle DBC.

Explanation of Solution

Given:

  $\begin{array}{l}\to \overline{AB}\text{ and }\overline{BC}\text{ are tangent}\\ \to AD>DC\end{array}$

  

Concept:

TheConverse 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:

  $\begin{array}{l}\text{⇒}AB=BC\left(\begin{array}{l}\text{Two tangent segments to a circle share a common endpoint outside}\\ \text{ the circle,then two segmets are congruent}\end{array}\right)\\ \text{⇒}DB=DB\left(\text{Reflexive property of congruence}\right)\\ \text{⇒}AD>DC\left(\text{Given}\right)\\ \text{⇒}\text{Conclusion : }\angle ABD>\angle DBC\left(\text{SSS inequaliy theorem}\right)\end{array}$

Conclusion:

Thus, it is proved that $\angle ABD>\angle DBC$ .