9. Let $A B C$ be a triangle, and let $B C D E, C A F G, A B H I$ be squares that do not overlap the triangle with centers $X, Y, Z$ respectively. Given that $A X=6, B Y=7$, and $C Z=8$, find the area of triangle $X Y Z$.
Answer:
$$
\frac{21 \sqrt{15}}{4}
$$

By the degenerate case of Von Aubel's Theorem we have that $Y Z=A X=6$ and $Z X=B Y=7$ and $X Y=C Z=8$ so it suffices to find the area of a $6-7-8$ triangle which is given by $\frac{21 \sqrt{15}}{4}$.
To prove that $A X=Y Z$, note that by LoC we get
$$
Y X^{2}=\frac{b^{2}}{2}+\frac{c^{2}}{2}+b c \sin \angle A
$$
and
$$
\begin{aligned}
A X^{2} & =b^{2}+\frac{a^{2}}{2}-a b(\cos \angle C-\sin \angle C) \\
& =c^{2}+\frac{a^{2}}{2}-a c(\cos \angle B-\sin \angle B) \\
& =\frac{b^{2}+c^{2}+a(b \sin \angle C+c \sin \angle B)}{2} \\
& =\frac{b^{2}}{2}+\frac{c^{2}}{2}+a h
\end{aligned}
$$
where $h$ is the length of the $A$-altitude of triangle $A B C$. In these calculations we used the well-known fact that $b \cos \angle C+c \cos \angle B=a$ which can be easily seen by drawing in the $A$-altitude. Then since $b c \sin \angle A$ and $a h$ both equal twice the area of triangle $A B C$, we are done.