Chapter 15.3, Problem 23PSC

To determine

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

Explanation of Solution

Given:

  $\begin{array}{l}\Delta BCD\text{ is isosceles with base }\overline{CD}\\ \angle ABD>\angle ABC\end{array}$

  

Formula 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:

  

  $\begin{array}{l}\to \text{In }\Delta ABC\text{ and }\Delta ABD\\ \text{⇒}BC=BD\\ \text{⇒}AB=AB\\ \text{⇒}\angle ABD>\angle ABC\\ \text{⇒}AD>AC\\ \text{⇒}\text{In }\Delta ACD\\ \text{⇒}\angle ACD>\angle ADC\end{array}$

Conclusion:

Thus from the calculation it is proved that $\angle ACD>\angle ADC$