Similar Triangles Problem For Geometry I In quadrilateral QRTS, we have QR = 11, QS = 9, and ST=2. Sides RQ and ST are extended past Q and S, respectively, to meet at point P. If PS = 8 and PQ = 5, then what is RT?
Try drawing it out. You should end up with triangle PTR, with point S lying on PT and Q lying on PR. You also know the length of PT and PR, as well as that of SQ.

In triangle PSQ, we have all the side lengths so you can find the angle between TP and PR using the law of cosines. 

Now you have, in PTR, two side lengths and one angle, so you can then use the law of cosines once again to find TR.
you can solve it without cosines by flipping the PQS triangle, so that you can use SAS similarity to prove they are similar.
PQ =5  PT = 10
PS = 8 PR = 16
and the angle between them is the same angle, so the triangles are similar and have a 2:1 ratio
so 9*2 = 18 and youre done.