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# 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.