Chapter 6.1, Problem 13WE

To determine

To Prove: $m\angle K>m\angle C$ shown in figure such that point $K$ lies inside $\Delta ABC$ .

Explanation of Solution

Given:

Figure is shown as.

  

point $K$ lies inside $\Delta ABC$ ..

Calculation:

Using two-column form.

  $\begin{array}{cc}\text{Statements}& \text{Reasons}\\ \begin{array}{l}\angle C+\angle CAB+\angle CBA=180\\ \angle KAB+\angle K+\angle KBA=180\end{array}& \text{Sum of angles in a }\Delta \text{ is 180}\\ \angle C+\angle CAB+\angle CBA=\angle KAB+\angle K+\angle KBA& \text{Using Transitive Property}\\ \begin{array}{l}\angle CAB=\angle CAK+\angle KAB,\\ \angle CBA=\angle CBK+\angle KBA,\end{array}& \text{A whole is equal to sum of its parts}\\ \begin{array}{l}\angle C+\angle CAK+\angle KAB+\angle CBK+\angle KBA=\angle KAB+\angle K+\angle KBA\\ \angle C+\angle CAK+\angle CBK=\angle K\\ m\angle K>m\angle C\end{array}& \begin{array}{l}\text{Substituting}\\ \angle CAB\text{and}\angle CBA\\ \text{A whole is greater than any of its part }\end{array}\end{array}$

Therefore, it has been proved.