20th Turkish Mathematical Olympiad

First observe that $\angle PAR=\angle QBR$. Let $\angle PAR=\alpha$, $\angle PBC=\beta$ and $\angle BAD=\theta$. Applying trigonometric form of Ceva Theorem for the point $P$ inside the triangle $ABC$ gives $$\frac{\sin (\theta +\alpha )}{\sin (\theta -\alpha )}\cdot \frac{\cos (\alpha +\beta +\theta )}{\sin (\alpha +\beta )}\cdot \frac{\sin \beta }{\cos (\beta +\theta )}=1(1).$$ Recall that $2\cos (\alpha +\beta +\theta )\sin \beta =\sin (\alpha +2\beta +\theta )-\sin (\alpha +\theta )$ and $2\sin (\alpha +\beta )\cos (\beta +\theta )=\sin (\alpha +2\beta +\theta )-\sin (\theta -\alpha )$. Thus, we can rewrite (1) as $$\frac{\sin (\theta +\alpha )}{\sin (\theta -\alpha )}\cdot \frac{\sin (\alpha +2\beta +\theta )-\sin (\alpha +\theta )}{\sin (\alpha +2\beta +\theta )-\sin (\theta -\alpha )}=1,$$ that is $(\sin (\theta +\alpha )-\sin (\theta -\alpha ))\sin (\alpha +2\beta +\theta )={\sin }^2(\theta +\alpha )-{\sin }^2(\theta -\alpha )$. Then we obtain $$\sin (\alpha +2\beta +\theta )=\sin (\theta +\alpha )+\sin (\theta -\alpha )=2\sin \theta \cos \alpha (2).$$

On the other hand applying trigonometric form of Ceva Theorem for the point $R$ inside the triangle $ATS$ results $$\frac{\sin (\alpha +2\beta +\theta )}{\sin \theta }\cdot \frac{\sin \alpha }{\sin (2\alpha )}\cdot \frac{\sin \angle RST}{\sin \angle RTS}=1(3).$$ Since $\sin (2\alpha )=2\cos \alpha \sin \alpha$ and $\sin (\alpha +2\beta +\theta )=2\sin \theta \cos \alpha$ by (2), (3) can be rewritten as $$\frac{2\sin \theta \cos \alpha }{\sin \theta }\cdot \frac{1}{2\cos \alpha }\cdot \frac{\sin \angle RST}{\sin \angle RTS}=1.$$ Finally we get $\sin \angle RST=\sin \angle RTS$ and the result follows.