Square $A B C D$ has sides of length 2. The midpoint of $B C$ is $E$. Point $F$ is on $A E$ so that $D F$ is perpendicular to $A E$. Determine the length of $D F$.

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Since $A B C D$ is a square, $\angle F A D+\angle F A B=90^{\circ}$.

Since $\triangle B A E$ is right-angled, we also have that $\angle B E A+\angle E A B=90^{\circ}$.

Since $\angle F A B=\angle E A B$, we then have $\angle F A D+\angle E A B=\angle B E A+\angle E A B$, from which it follows that $\angle F A D=\angle B E A$.

Two right-angled triangles are similar if they have a non-right angle in common, so $\triangle F A D$ is
similar to $\triangle B E A$.

This means $\frac{F D}{B A}=\frac{A D}{E A}$ so

$$
F D=\frac{B A \times A D}{E A}=\frac{2 \times 2}{E A}=\frac{4}{E A} .
$$

Since $E$ is the midpoint of $B C$ and $B C=2$, we have $B E=1$.

By the Pythagorean theorem, $E A^{2}=B A^{2}+B E^{2}=2^{2}+1^{2}=5$.

Since $E A>0$, we have $E A=\sqrt{5}$ and thus $F D=\frac{4}{\sqrt{5}}$.

ANSWER: $\frac{4}{\sqrt{5}}$