Chapter 15.1, Problem 13PSB

To determine

To find and justify: the relation between $\angle DBC\text{ and }\angle DCB$ .

Answer to Problem 13PSB

  $XV>VY$ .

Explanation of Solution

Given:

  

In the given figure,

  $\begin{array}{l}\angle ABC>\angle ACB\\ \overrightarrow{BD}\text{ bisects}\angle ABC\\ \overrightarrow{CD}\text{ bisects}\angle ACB\end{array}$ .

Calculation:

According to the question,

  $\begin{array}{l}\angle DBC=\frac{1}{2}\angle ABC\left\{\text{angle-bisctors}\right\}\\ \angle DCB\cong \frac{1}{2}\angle ACB\left\{\text{angle-bisctors}\right\}\\ \angle ABC>\angle ACB\\ \Rightarrow \frac{1}{2}\angle ABC>\frac{1}{2}\angle ACB\\ \angle DBC>\angle DCB\end{array}$

Hence, $\angle DBC>\angle DCB$