Question

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

Answer 1

Given, Bisectors of $\angle PQR$and $\angle PRS$ meet at point $T$.

To prove: $\angle QTR=\frac{1}{2}\angle QPR$.

Proof,

$\angle TRS=\angle TQR+\angle QTR$ (Exterior angle of a triangle equals to the sum of the two interior angles.)

$\Rightarrow \angle QTR=\angle TRS-\angle TQR$ --- (i)

Also $\angle SRP=\angle QPR+\angle PQR$

$2\angle TRS=\angle QPR+2\angle TQR$

$\angle QPR=2\angle TRS-2\angle TQR$

$\Rightarrow \frac{1}{2}\angle QPR=\angle TRS-\angle TQR$ --- (ii)

Equating (i) and (ii),

$\therefore \angle QTR=\frac{1}{2}\angle QPR$ $\left[henceproved\right]$