Chapter 4.7, Problem 24WE

To determine

To prove: $\overrightarrow{BP}$ bisects $\angle ABC$ .

Explanation of Solution

Given information:

  $\overrightarrow{DP}$ bisects $\angle ADE$ ;

  $\overrightarrow{EP}$ bisects $\angle DEC$

Given figure,

  

Calculation:

Line $\overrightarrow{DP}$ bisects $\angle ADE$ , therefore point $P$ is equidistant from sides $\overline{AD}$ and $\overline{DE}$ .

Also $\overrightarrow{EP}$ bisects $\angle DEC$ , therefore point $P$ is equidistant from sides $\overline{DE}$ and $\overline{CE}$ .

Point $P$ is equidistant from sides $\overline{AD}$ and $\overline{DE}$ and also $\overline{DE}$ and $\overline{CE}$ .

Therefore, point $P$ is equidistant from sides $\overline{AD}$ and $\overline{CE}$ , or one can say that point $P$ is equidistant from sides $\overline{AB}$ and $\overline{CB}$ .

Since point $P$ is equidistant from sides $\overline{AB}$ and $\overline{CB}$ , then , point $P$ lies on the bisector of $\angle ABC$ or $\overrightarrow{BP}$ bisects $\angle ABC$ .

Hence proved