# Find length of a segment in a trapezoid I tried a lot of similarity between triangles, because ACQP and ACST are trapezoids, so i can work with a lot of proportional segments in some of the triangles, but i found nothing.

Any hints?

• It seems that there is not a unique solution - for every selection of Q you could move C MS=1 (At least it seems so) - may be you are missing a piece of data? – Moti Jul 12 at 7:08
• How Q and C could move if they are the vertexes? – Rodrigo Pizarro Jul 12 at 7:10
• They are not defined as fixed (unless you "forgot" to mention that this is isosceles trapezoid). – Moti Jul 12 at 7:13
• I dont think there is a problem. Yes, you can move them around and it will change a lot of the lengths. But the interesting property here is AM, which will stay the same. – Nurator Jul 12 at 8:25

The solution is as you expected the usage of similar triangles. The first thing you need to notice is that the triangles TMS and MAC are similar, thus allowing us to write $$\frac{AM}{MS}=\frac{AC}{TS}$$
Then note that QTS and QAC are similar, thus $$\frac{AC}{TS}=\frac{QS+SC}{QS}$$
Lastly, lengthen the lins PA and QC to get an intersection point O, building the triangle OAC. With this, we can see that $$\frac{QS+SC}{QS}=\frac{PR+AR}{PR}$$
Altogether, we can write $$\frac{AM}{MS}=\frac{PR+AR}{PR}$$
with only one unknown. Inserting AR=4, PR=2 and MS=1, we get $$\frac{AM}{1}=\frac{2+4}{2} \Rightarrow AM=3$$
$$AR:PR=2:1=SC:QS$$ (Thales proportional segments theorem). Thus, $$QC=3QS$$. It follows from similarity that $$QC:QS=AC:TS=3:1$$. Again, by similarity we see that $$AM:MS=3:1$$.