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


Question

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:

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

Reasoning:

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.

Steps:

Given,

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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