7. In the triangle $A B C, A B=8, B C=7$ and $C A=6$. Let $E$ be the point on $B C$ such that $\angle B A E=3 \angle E A C$. Find $4 A E^{2}$.
7. Answer: 135 .
In general, we let $A B=c, A C=b$ and $B C=a$. Let $A D$ be the bisector of $\angle A$. Using the angle-bisector theorem, $B D / C D=c / b$. Thus $B D=a c /(b+c)$ and $C D=a b /(b+c)$. By Stewart's theorem.
$$
\left(\frac{a b}{b+c}\right) c^{2}+\left(\frac{a c}{b+c}\right) b^{2}=a A D^{2}+\frac{a b c}{(b+c)^{2}} a^{2},
$$
which after solving for $A D^{2}$ gives
$$
A D^{2}=b c\left[1-\left(\frac{a}{b+c}\right)^{2}\right] .
$$

Substituting the values of $a=7, b=6, c=8$, we get $A D=6$. Thus $\triangle A C D$ is isosceles and the bisector $A E$ of $\angle C A D$ is perpendicular to $B C$ and $E$ is the midpoint of $C D$. As $D C=3$, we have $E C=3 / 2$. Using Pythagoras' theorem, we get $A E^{2}=A C^{2}-E C^{2}=$ $6^{2}-\left(\frac{3}{2}\right)^{2}=135 / 4$. Therefore, $4 A E^{2}=135$.