"A circle of radius $5$ is inscribed in an isosceles trapezoid $ABCD$. If the major base measures $20$, what is the distance between the tangency points of the non-parallel sides?"
My first approach was defining a point $F \in CD$ such that $AF \perp DC$.
Then, for an isosceles trapezium to have an inscribed circle, it must satisfy that $AB+CD=AD+BC$
Define $AD=BC=y$ and $AB=x$
Then: $20+x=2y$.
$AF=10$
By Pythagoras' theorem: $DF^2+AF^2=AD^2$ = $(\frac {20-x}{2})^2+10^2=y^2$
Solving this equation system I got that $x=5, y=\frac{25}{2}$, but I don't see how to continue. I think that calculating $MN$ solves the problem but I don't know how.
Any hints? Are my approaches wrong?
You're basically done. Since $AM = \frac12AB$ by external tangents, you know the ratio $\frac{AM}{AD}$, and $\frac{MD}{AD}$, so you can take a weighted average of $AB$ and $DC$ to get $MN$.
(If you prefer, you can draw $X$ the intersection of $AD$ and $BC$ and use the similar triangles $\triangle ABX$, $\triangle MNX$, $\triangle DCX$, which is an equivalent approach.)