Ex.6.3 Q6 Lines and Angles Solution - NCERT Maths Class 9


In Fig. below, the side \(QR\) of \(\Delta PQR\) is produced to a point \(S\). If the bisectors of \(\angle PQR\) and \(\angle PRS\) meet at point \(T\), then prove that \(\begin{align} \angle QTR = \frac{1}{2}\angle QPR\end{align} \) .

Text Solution

What is known?

The side \(QR\) of \(\Delta PQR\) is produced to a point \(S\) and the bisectors of \(\angle PQR\) and \(\angle PRS\) meet at point \(T\).

To prove:

\(\begin{align}\angle QTR = \frac{1}{2}\angle QPR\end{align}\)


As we know exterior angle of a triangle:

If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.



Bisectors of \(\angle PQR\) and \(\angle PRS\) meet at point \(T\).

Hence, \(TR\) is a bisector of \(\angle PRS\) and \(TQ\) is a bisector of \(\angle PQR\) 

\(\begin{align} \angle PRS &= 2\angle TRS\quad\dots\left( {\rm{i}} \right)\\\angle PQR &= 2\angle TQR\quad\dots\left( {{\rm{ii}}} \right)\end{align} \)

Now, in \(\Delta TQR\)

\(\begin{align}\angle TRS&= \angle TQR + \angle QTR\\ &\left( {{\text{Exterior angle of triangle}}} \right)\\\\\angle QTR &= \angle TRS-\angle TQR\quad\dots\left( {{\text{iii}}} \right)\end{align}\)

Similarly, in \(\Delta PQR\)

\(\begin{align}\angle PRS &= \angle QPR + \angle PQR\qquad\\ &\left( {{\text{Exterior angle of triangle}}} \right)\\\\2\angle TRS &= \angle QPR + 2\angle TQR\\ &\left[ {{\text{From }}\left( {\rm{i}} \right){\rm{ and }}\left( {{\rm{ii}}} \right)} \right]\\\\\angle QPR &= 2\angle TRS-2\angle TQR\\\angle QPR &= 2\left( {\angle TRS-\angle TQR} \right)\\\angle QPR &= 2\angle QTR\qquad\qquad\,\,\,\,\,\qquad\\ &\left[ {{\text{From }}\left( {{\rm{iii}}} \right)} \right]\\\\\angle QTR &= \frac{1}{2}\angle QPR\end{align}\)

Hence proved.

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