Prove that $\angle QPR=\frac12\angle QTR$ without using trigonometry

Question

The bisectors of angle $Q$ and angle $R$ meet at $T$.

What would be the shortest method to prove $\angle QPR=\frac12\angle QTR$?

Answer 1

Is that result even right?

Let $\angle QTR = a \:\: \angle QPR = b$ $$\frac{1}{2} \angle R = \frac{1}{2} \angle Q + a \: \:\text {(external angles)}$$ $$\angle R = \angle Q + b$$ $$\therefore b = 2a$$

Answer 2

$$\begin{align*} \angle QTR&=\pi-\angle TQR-\angle TRQ \\ &=\pi-\frac12\angle PQR-(\frac{\pi}{2}+\frac12\angle PRQ)\\ &=\frac{\pi}{2}-\frac12\angle PQR-\frac12\angle PRQ\\ &=\frac{1}{2}(\pi-\angle PQR-\angle PRQ) \end{align*}$$

Can you finish from here?

Answer 3

What you want to prove is actually $\angle QPR = 2 \, \angle QTR$. You do not need trigonometry for this. Just let $\angle PQR = \alpha$ and $\angle QRP = \beta$. The external angle at vertex $R$ is $\pi-\beta$. Now you know the angles at vertices $Q$ and $R$ of triangles $QRP$ and you can immediately find the angles at vertices $Q$ and $R$ of triangle $QRT$, knowing that $QT$ and $RT$ are angle bisectors. Finally, to find the angles at vertices $P$ and $T$ you just need to use the fact that the sum of the three interior angles of a triangle is always $\pi.$