Question

In the following figure the side QR of Δ PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at a point T, then prove that ∠QTR = $\frac{1}{2}$∠QPR.

Answer 1

From the figure,
$$\begin{gathered} \angle PQT=\angle TQR=\frac{1}{2}\angle PQR.....\left(i\right) \\ \angle PRT=\angle TRS=\frac{1}{2}\angle PRS.....\left(ii\right) \end{gathered}$$
Here, $\angle PRS$ is an external angle for $∆PQR$. So
$$\begin{gathered} \angle PRS=\angle PQR+\angle QPR \\ \Rightarrow 2\angle TRS=2\angle TQR+\angle QPR.....\left(iii\right)\left[\mathrm{From}\left(i\right)\mathrm{and}\left(ii\right)\right] \end{gathered}$$
$\angle TRS$ is an external angle for $∆TQR$. So
$\angle TRS=\angle QTR+\angle TQR$
Substituting $\angle TRS=\angle QTR+\angle TQR$ in (iii), we get
$$\begin{gathered} 2\left(\angle QTR+\angle TQR\right)=2\angle TQR+\angle QPR \\ \Rightarrow 2\angle QTR+2\angle TQR=2\angle TQR+\angle QPR \\ \Rightarrow 2\angle QTR=\angle QPR \\ \Rightarrow \angle QTR=\frac{1}{2}\angle QPR \end{gathered}$$
Hence, $\angle QTR=\frac{1}{2}\angle QPR$.