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Answer :

In \(\triangle{PQS}\), we have,

\(\angle{QPR}\) + \(\angle{PQR}\) = \(\angle{PRS}\) ...(i)

As we know, Sum of interior opposite angles is equal to exterior angle.

Similarly in \(\triangle{TQR}\), we have,

\(\Rightarrow \) \(\angle{QTR}\) + \(\angle{TQR}\) = \(\angle{TRS}\) ...(ii)

As, QT and RT are the bisectors of \(\angle{PRS}\) and \(\angle{PQR}\), respectively, we get,

\(\Rightarrow \) \(\angle{TRS}\) = (\(\frac{1}{2} \) ) \(\angle{PRS}\) ...(iii)

\(\Rightarrow \) \(\angle{TQR}\) = (\(\frac{1}{2} \) ) \(\angle{PQR}\) ...(iv)

By multiplying both sides of eq.(i), we get,

(\(\frac{1}{2} \) )\(\angle{QPR}\) + (\(\frac{1}{2} \) )\(\angle{PQR}\) = (\(\frac{1}{2} \) )\(\angle{PRS}\)

from eq. (iii) and (iv),

(\(\frac{1}{2} \) )\(\angle{QPR}\) + \(\angle{TQR}\) = \(\angle{TRS}\) ...(v)

Thus, from eq.(ii) and (v), we get,

\(\Rightarrow \) \(\angle{QTR}\) + \(\angle{TQR}\) = (\(\frac{1}{2} \) ) \(\angle{QPR}\) + \(\angle{TQR}\)

cancelling equal terms, we get,

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

Hence, proved.

- In figure, sides QP and RQ of \(\angle{PQR}\) are produced to points S and T, respectively. If \(\angle{SPR}\) = \(135^\circ\) and \(\angle{PQT}\) = \(110^\circ\) , find \(\angle{PRQ}\).
- In figure, \(\angle{X}\) = \(60^\circ\) , \(\angle{XYZ}\) = \(54^\circ\) , if YO and ZO are the bisectors of \(\angle{XYZ}\) and \(\angle{XZY}\) respectively of \(\triangle{XYZ}\) , find \(\angle{OZY}\) and \(\angle{YOZ}\).
- In figure, if AB || DE , \(\angle{BAC}\) = \(35^\circ\) and \(\angle{CDE}\) = \(53^\circ\) , find \(\angle{DCE}\).
- In figure, if lines PQ and RS interest at point T, such that \(\angle{PRT}\) = \(40^\circ\) ,\(\angle{RPT}\) = \(95^\circ\) and \(\angle{TSQ}\) = \(75^\circ\) , find \(\angle{SQT}\).
- In figure, if PQ is perpendicular to PS , PQ || SR , \(\angle{SQR}\) = \(28^\circ\) and \(\angle{QRT}\) = \(65^\circ\) , then find the values of x and y.

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