In figure, the side QR of \(\angle{PQR}\) is produced to a point S. If the bisectors of \(\angle{PQR}\) and \(\angle{PRS}\) meet at point T, then prove that \(\angle{QTR}\) = (1/2) \(\angle{QPR}\)

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.