Chapter 1.5, Problem 14PSA

To determine

To Prove: $\overrightarrow{EB}\text{ and }\overrightarrow{EC}\text{trisect}\angle AED$

Explanation of Solution

Given:

  

  $\angle AEB\cong \angle BEC\cong \angle CED$

Concept Used:

Trisector of an angle divides into three parts.

Proof:

Here, we have

  $\begin{array}{l}\angle AEB\cong \angle BEC\cong \angle CED\\ \Rightarrow m\angle AEB=m\angle BEC=m\angle CED\\ \Rightarrow \overrightarrow{EB}\text{ and }\overrightarrow{EC}\text{trisects }\angle AED\end{array}$