1. In triangle $A B C, A B=6, B C=10$, and $C A=14$. If $D, E$, and $F$ are the midpoints of sides $B C$, $C A$, and $A B$, respectively, find $A D^{2}+B E^{2}+C F^{2}$.
Answer: 249
Solution: We use Stewart's Theorem with the median $A D$ : (Note that $B D=C D=\frac{B C}{2}$.)
$$
C A^{2}(B D)+A B^{2}(C D)=B C\left(A D^{2}+(B D)(C D)\right) \rightarrow \frac{1}{2}\left(C A^{2}+A B^{2}\right)=A D^{2}+\left(\frac{B C}{2}\right)^{2}
$$

Similarly, by using Stewart's Theorem with the medians $B E$ and $C F$, we get:
$$
\frac{1}{2}\left(B C^{2}+A B^{2}\right)=B E^{2}+\left(\frac{C A}{2}\right)^{2}
$$
and
$$
\frac{1}{2}\left(C A^{2}+B C^{2}\right)=C F^{2}+\left(\frac{A B}{2}\right)^{2} .
$$

Adding these three equations, we have:
$$
A B^{2}+B C^{2}+C A^{2}=A D^{2}+B E^{2}+C F^{2}+\frac{1}{4}\left(A B^{2}+B C^{2}+C A^{2}\right) .
$$

Thus,
$$
A D^{2}+B E^{2}+C F^{2}=\frac{3}{4}\left(A B^{2}+B C^{2}+C A^{2}\right)=\frac{3}{4}\left(6^{2}+10^{2}+14^{2}\right)=249 .
$$