
cyclic quadrilateral
PQRS is a cyclic quadrilateral where angle PQR and angle QRS are obtuse and have the same value . IF PQ=9 , QR=6 , and RS=4 , prove that if T is the point of intersection of the diagonals , then the circle that passes through R,S and T touches QR at R .

I think it's something wrong with this problem.
If $\displaystyle \widehat{PQR}=\widehat{QRS}=\alpha$ then $\displaystyle \widehat{QPS}=\widehat{RSP}=180\alpha$
Then $\displaystyle \widehat{QRS}+\widehat{RSP}=180\Rightarrow QR\parallel PS$ and the quadrilater is an isosceles trapezoid. Then $\displaystyle PQ=RS$.
But $\displaystyle PQ=9, \ RS=4$.

If this inscribed quadrilateral has 2 equal obtuse angles, it is most certainly an isosceles trapezoid / trapezium, and so PQ and RS is supposed to have the same length.