Chapter 15, Problem 14RP

To determine

To Prove: In the given figure one of the segments is greater than the other.

Explanation of Solution

Given:

  $ED>EB$

  

  $\begin{array}{l}AB<AD\\ \overrightarrow{BE}\text{ bisects }\angle \text{ABC}\\ \overrightarrow{DE}\text{ bisects }\angle \text{ADC}\end{array}$

Formula Used:

In a triangle angle opposite to larger side is always larger, and smallest side is smallest.

Calculation:

  $\begin{array}{l}\text{In }\Delta ABD\\ AB<AD\\ \text{⇒}\angle D<\angle B\\ AB<AD\\ 2y<2x\\ ory<x\\ \text{In }\Delta \text{BED}\\ \text{larger angles corresponds to larger sides}\\ ED>EB\\ \text{Hence Proved}\text{.}\end{array}$

Conclusion:

Hence, $\text{Proved}\text{.}$