# Find angle QTS.

anglegeometrytriangles

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.

Let add one point $$O$$ to construction, which is intersection of $$SU$$ and $$PQ$$.
$$\angle TPS=\angle PRQ=\angle PQR=\angle OPS \Rightarrow TS=OS\Rightarrow OT=2OS$$
$$\frac{PQ}{SU}=\frac{OP}{OS} \Rightarrow \frac{PQ}{QT}=\frac12 \frac{PQ}{SU}=\frac12\frac{OP}{OS}=\frac{OP}{OT}\Rightarrow \frac{PQ}{OP}=\frac{QT}{OT} \Rightarrow TP {\rm\ is\ bisector\ of\ }\angle QTO$$
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 $$2\alpha+\alpha+2\alpha-90°=180°$$, $$5\alpha=270°$$, $$\alpha=54°$$. $$\angle QTS=2\alpha=108°$$.