Chapter 15.3, Problem 18PSB

To determine

To Prove: The segment PQ is greater than segment PRin the given figure.

Explanation of Solution

Given:

  $\begin{array}{l}\to QT>TR\\ \to \overrightarrow{QT}\text{ and }\overrightarrow{QS}\text{ bisect }\angle PQR\\ \to \overrightarrow{RS}\text{ and }\overrightarrow{RT}\text{ trisect }\angle PRQ\end{array}$

  

Formula Used:

The Hinge Theorem: It states that if two sides of two triangles are congruent and angle of one is greater than the angle of second then the third side of the triangle will be greater than the another.

Calculation:

  

  $\begin{array}{l}\text{⇒}QT>TR\left(\text{Given}\right)\\ \text{⇒}\text{In }\Delta QTR,x>y\left(\text{From the figure}\right)\\ \text{⇒}3x>3y,\angle R>\angle Q\\ \text{⇒}PQ>PR\end{array}$

Conclusion:

Thus from the calculation it is proved that $PQ>PR$