5. A square $A B C D$ has side-length 2 , and $M$ is the midpoint of $B C$. The circle $S$ inside the quadrilateral $A M C D$ touches the three sides $A M, C D$ and $D A$. What is its radius?
In any geometry question, drawing a diagram is an important first task:

Let $O$ be the centre of circle $S$.
Construct the perpendicular to $A B$ through $O$, and denote the points where it intersects $A B, A M$ and $C D$ as $P, Q$ and $U$ respectively. Let $T$ denote the point where $A M$ meets the circle, and let $r$ denote the radius of the circle.

Note that $A P=r$, and also $A M=\sqrt{5}$ by Pythagoras' Theorem.
Then $\angle O T Q=90^{\circ}$ (tangent and radius),
$\angle O Q T=\angle B M Q$ (alternate angles).
Hence $\triangle O Q T$ is similar to $\triangle A M B$. Hence $\frac{O Q}{O T}=\frac{A M}{A B}=\frac{\sqrt{5}}{2}$, so $O Q=\frac{\sqrt{5}}{2} r$.
Also, $\triangle A Q P$ is similar to $\triangle A M B$ (since $P Q$ is parallel to $B M$ ), so $\frac{P Q}{A P}=\frac{B M}{A B}=\frac{1}{2}$, so $P Q=\frac{1}{2} A P=\frac{r}{2}$.
Then
$$
2=P Q+Q O+O U=\frac{1}{2} r+\frac{\sqrt{5}}{2} r+r=\frac{3+\sqrt{5}}{2} r
$$

So
$$
r=\frac{2}{\frac{3+\sqrt{5}}{2}}=\frac{4}{3+\sqrt{5}}=3-\sqrt{5} .
$$