Question

$S$ is a point on the side $QR$ of triangle $PQR$ such that $PS$ is the bisector of angle $QPR$ then:

• A. $QS=SR$
• B. $QP>QS$
• C. $QS>QP$
• D. $SR>RP$

Answer 1

The correct option is B

$QP>QS$

According to the given condition, $\angle QPS=\angle RPS$

By external angle theorem,

$$\begin{gathered} \angle PSQ=\angle RPS+\angle PRS \\ \Rightarrow \angle PSQ=\angle QPS+\angle PRS \\ \Rightarrow \angle PSQ>\angle QPS \end{gathered}$$

In a triangle, the side opposite to the greater angle is greater. Therefore,

$\Rightarrow QP>QS$

Hence, option B is correct.