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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.