Q

# In Fig. 6.44, the side QR of ∆ PQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, then prove that ∠ QTR = 1 / 2 ∠ QPR.

6. In Fig. 6.44, the side QR of $\Delta$ 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 = $\frac{1}{2}$   $\angle$QPR.

Views

We have,
$\Delta$ PQR is produced to a point S and  bisectors of $\angle$ PQR and $\angle$ PRS meet at point T,
By exterior angle sum property,
$\angle$ PRS = $\angle$P + $\angle$PQR
Now, $\frac{1}{2}\angle PRS =\frac{1}{2}\angle P + \frac{1}{2} \angle PQR$
$\Rightarrow \angle TRS = \frac{1}{2}\angle P + \angle TQR$................(i)

Since QT and QR are the bisectors of  $\angle$ PQR and $\angle$ PRS respectively.

Now, in $\Delta$QRT,
$\angle TRS = \angle T + \angle TQR$..............(ii)

From eq (i) and eq (ii),  we get

$\frac{1}{2}\angle P= \angle T$
$\Rightarrow \frac{1}{2}\angle QPR= \angle QTR$
Hence proved

Exams
Articles
Questions