5. Let $A B C D$ be a convex trapezoid such that $\angle D A B=\angle A B C=90^{\circ}, D A=2, A B=3$, and $B C=8$. Let $\omega$ be a circle passing through $A$ and tangent to segment $\overline{C D}$ at point $T$. Suppose that the center of $\omega$ lies on line $B C$. Compute $C T$.
Answer:
$4 \sqrt{5}-\sqrt{7}$

Solution:
Let $A^{\prime}$ be the reflection of $A$ across $B C$, and let $P=A B \cap C D$. Then since the center of $\omega$ lies on $B C$, we have that $\omega$ passes through $A^{\prime}$. Thus, by power of a point, $P T^{2}=P A \cdot P A^{\prime}$. By similar triangles, we have
$$
\frac{P A}{A D}=\frac{P B}{B C} \Longrightarrow \frac{P A}{2}=\frac{P A+3}{8} \Longrightarrow P A=1,
$$
and $A^{\prime} P=1+2 \cdot 3=7$, so $P T=\sqrt{7}$. But by the Pythagorean Theorem, $P C=\sqrt{P B^{2}+B C^{2}}=4 \sqrt{5}$, and since $T$ lies on segment $C D$, it lies between $C$ and $P$, so $C T=4 \sqrt{5}-\sqrt{7}$.