8. Let $A B T C D$ be a convex pentagon with area 22 such that $A B=C D$ and the circumcircles of triangles $T A B$ and $T C D$ are internally tangent. Given that $\angle A T D=90^{\circ}, \angle B T C=120^{\circ}, B T=4$, and $C T=5$, compute the area of triangle $T A D$.
Answer: $64(2-\sqrt{3})$
Solution: Paste $\triangle T C D$ outside the pentagon to get $\triangle A B X \cong \triangle D C T$. From the tangent circles condition, we get
$$
\begin{aligned}
\angle X B T & =360^{\circ}-\angle X B A-\angle A B T \\
& =360^{\circ}-\angle D C T-\angle A B T \\
& =360^{\circ}-270^{\circ}=90^{\circ} \\
\angle X A T & =90^{\circ}-\angle B X A-\angle A T B \\
& =90^{\circ}-\angle C T D-\angle A T B \\
& =90^{\circ}-\left(120^{\circ}-90^{\circ}\right)=60^{\circ}
\end{aligned}
$$

Moreover, if $x=A T$ and $y=T D$, then notice that
$$
\begin{aligned}
{[A B T C D] } & =[A B T]+[C D T]+[A T D] \\
& =[X A T]-[X B T]+[A T D] \\
& =\frac{1}{2} x y \sin 60^{\circ}-\frac{1}{2} \cdot 4 \cdot 5+\frac{1}{2} x y \\
& =\frac{2+\sqrt{3}}{4} x y-10
\end{aligned}
$$
so we have
$$
x y=32 \cdot \frac{4}{2+\sqrt{3}}=128(2-\sqrt{3}) \Longrightarrow[A T D]=\frac{1}{2} x y=64(2-\sqrt{3}) \text {. }
$$