Find length of a segment in a trapezoid

Question

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?

Answer 1

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 $$ \frac{AM}{1}=\frac{2+4}{2} \Rightarrow AM=3 $$

Answer 2

$AR:PR=2:1=SC:QS$ (Thales proportional segments theorem). Thus, $QC=3QS$. It follows from similarity that $QC:QS=AC:TS=3:1$. Again, by similarity we see that $AM:MS=3:1$.