4. In triangle $A B C$, we have $A B=B C=5$ and $C A=8$. What is the area of the region consisting of all points inside the triangle which are closer to $A B$ than to $A C$ ?
Answer: $\frac{60}{13}$
Solution: Note that $A B C$ is simply an isosceles triangle obtained by joining two 3-4-5 triangles; hence, its area is $3 \cdot 4=12$. Then, if $D$ is the foot of the angle bisector of $\angle A$ on $B C$, the area we want is actually just the area of $A B D$. By the angle bisector theorem,
$$
\frac{[A B D]}{[A D C]}=\frac{A B}{A C}=\frac{5}{8},
$$
and so the area of $A B D$ is $\frac{5}{13}$ that of $A B C$, that is, $\frac{5}{13} \cdot 12=\frac{60}{13}$.