Question

In a quadrilateral PQRS, if the bisectors of $\angle Q$ and $\angle R$ meet at T, then the value of $\angle QTR$ is always equal to

• A. $2(\angle P+\angle S)$
• B. $\frac{\angle P+\angle S}{2}$
  
• C. $\angle P+\angle S$
• D. $\frac{\angle P+\angle S}{4}$
  

Answer 1

The correct option is B $\frac{\angle P+\angle S}{2}$


Given that in a quadrilateral PQRS, the bisectors of ∠Q and ∠R meet at T.
$\therefore \angle QRT=\frac{1}{2}\angle R$ and $\angle RQT=\frac{1}{2}\angle Q$ ............(i)
Now, $\angle P+\angle Q+\angle R+\angle S={360}^0$ (Angle sum property of quadrilateral)
$\Rightarrow \angle Q+\angle R={360}^0–(\angle P+\angle S)$ …..(ii)
In $\Delta RQT$,
$\angle QRT+\angle RQT+\angle QTR={180}^0$ (Angle sum property of triangle)
$\Rightarrow \frac{1}{2}\angle R+\frac{1}{2}\angle Q+\angle QTR={180}^0$ ......(from (i))
$\Rightarrow \angle \text{QTR}={180}^0-\frac{1}{2}(\angle R+\angle Q)$
$\Rightarrow \angle \text{QTR}={180}^0-\frac{1}{2}[{360}^0-(\angle P+\angle S\left)\right]$ .........( from (ii))
$\Rightarrow \angle \text{QTR}={180}^0-{180}^0+\frac{1}{2}(\angle P+\angle S)$
$\Rightarrow \angle \text{QTR}=\frac{1}{2}(\angle P+\angle S)$
Hence, the correct answer is option (b).