3. $A B C D$ is a square and $X$ is a point on the side $D A$ such that the semicircle with diameter $C X$ touches the side $A B$. Find the ratio $A X: X D$.
Solution
Since we are interested in the ratio of sides, we can assign a side length for convenience as this would not affect the ratio of lengths. Let $C D=2$.

Let $O$ be the centre of the semicircle, let $Y$ be the point of tangency on $A B$ and let $Y O$ meet $C D$ at $Z$.

Note that $O Y=O X=O C$ as they are all radii. Since $O Y$ is perpendicular to $A B, O Z$ is perpendicular to $C D$. Then $Z$ is the midpoint of $D C$ because triangles $C O Z$ and $C X D$ are similar with scale factor 2 .
Let $X D=x$, so $O Z=\frac{x}{2}$ and $O Y=2-\frac{x}{2}$.
Hence $O C=2-\frac{x}{2}$ and, using Pythagoras' Theorem on triangle $C Z O$, we have $1+\left(\frac{x}{2}\right)^{2}=$ $\left(2-\frac{x}{2}\right)^{2}$
This leads to $x=\frac{3}{2}$, so $A X: X D=1: 3$.
Alternative
As in the previous solution, we show that $Y$ is the midpoint of $A B$.
$\angle X Y C=90^{\circ}$ because it is the angle in a semi-circle.
Then, $\angle C Y B=90^{\circ}-\angle X Y A=\angle A X Y$.
Hence $\triangle A X Y \sim \triangle B Y C$.
Therefore, $\frac{A X}{A Y}=\frac{B Y}{B C}=\frac{1}{2}$.
Now $A X=\frac{1}{2} A Y=\frac{1}{2} B Y=\frac{1}{4} B C=\frac{1}{4} A D$, so $A X: X D=1: 3$.