Find angle QTS.

Question

Given $PQ=PR, QT = 2SU$, $PS \parallel QR$ and $SU \perp QR$. $QT$ is angle bisector of $\angle PQR$. What is the size of $\angle QTS$?

What I have so far: $\angle PQT=\angle RQT=\alpha$, $\angle QRP = \angle SPR = 2\alpha$,$\angle QPT = 180-4\alpha$.

I am unsure how to proceed.

Answer 1

Let add one point $O$ to construction, which is intersection of $SU$ and $PQ$.

$$\angle TPS=\angle PRQ=\angle PQR=\angle OPS \Rightarrow TS=OS\Rightarrow OT=2OS$$

Using intercept theorem:

$$\frac{PQ}{SU}=\frac{OP}{OS} \Rightarrow \frac{PQ}{QT}=\frac12 \frac{PQ}{SU}=\frac12\frac{OP}{OS}=\frac{OP}{OT}\Rightarrow \frac{PQ}{OP}=\frac{QT}{OT} \Rightarrow TP {\rm\ is\ bisector\ of\ }\angle QTO$$

Let $\angle QOT=\alpha$, then $\angle PTO=\alpha$, $\angle QTS=2\alpha$, $\angle QTU=180°-2\alpha$, $\angle UQT=2\alpha-90°$, $\angle OQT=2\alpha-90°$. Then in triangle $OQT$ sum of angles is $2\alpha+\alpha+2\alpha-90°=180°$, $5\alpha=270°$, $\alpha=54°$. $\angle QTS=2\alpha=108°$.

Answer 2

Following on from where you left off:

You have $\angle PST=90^o$ and $\angle QTS=90^o+\alpha$

Applying the Sine Rule in $\triangle QPT$, $$\frac{QT}{\sin(180^o-4\alpha)}=\frac{PT}{\sin\alpha}\implies PT=\frac{QT\sin\alpha}{\sin4\alpha}$$

In $\triangle QUT$ we have $$UT=QT\sin\alpha$$

Now since $\angle SPR=2\alpha$, $$ST=PT\sin2\alpha=\frac{QT\sin\alpha\sin2\alpha}{\sin4\alpha}$$

Then, $$SU=ST+UT=QT\sin\alpha\left[1+\frac{\sin2\alpha}{\sin4\alpha}\right]=\frac12QT$$ $$\implies \sin\alpha\left[1+\frac{1}{2\cos2\alpha}\right]=\frac12$$

Using $\cos2\alpha=1-2\sin^2\alpha$ and simplifying leads to the cubic $$4\sin^3\alpha-2\sin^2\alpha-3\sin\alpha+1=0$$ $$\implies(\sin\alpha-1)(4\sin^2\alpha+2\sin\alpha-1)=0$$

Since $\alpha$ must be acute, the only solution from this is $\alpha=18^o$, and so the required $\angle QTS=108^o$.

However, due to the harmonious nature of the answer I am left with a nagging suspicion that there must be a better, more elegant way of obtaining this, using geometry, perhaps...