20. [11] Let $\triangle A B C$ be an isosceles right triangle with $A B=A C=10$. Let $M$ be the midpoint of $B C$ and $N$ the midpoint of $B M$. Let $A N$ hit the circumcircle of $\triangle A B C$ again at $T$. Compute the area of $\triangle T B C$.
Answer:
30
Solution: Note that since quadrilateral $B A C T$ is cyclic, we have
$$
\angle B T A=\angle B C A=45^{\circ}=\angle C B A=\angle C T A .
$$

Hence, $T A$ bisects $\angle B T C$, and $\angle B T C=90^{\circ}$. By the angle bisector theorem, we then have
$$
\frac{B T}{T C}=\frac{B N}{N C}=\frac{1}{3} .
$$

By the Pythagorean theorem on right triangles $\triangle T B C$ and $\triangle A B C$, we have
$$
10 B T^{2}=B T^{2}+T C^{2}=A B^{2}+A C^{2}=200,
$$
so $B T^{2}=20$. Note that the area of $\triangle T B C$ is
$$
\frac{B T \cdot T C}{2}=\frac{3 \cdot B T^{2}}{2},
$$
so our answer is then
$$
\frac{3}{2} \cdot B T^{2}=\frac{3}{2} \cdot 20=30
$$