6.3. The median of a tetrahedron is defined as the segment connecting its vertex to the point of intersection of the medians of the opposite face. Express the length of the median of a tetrahedron in terms of the lengths of its edges.
6.3. Let $A D=a, B D=b, C D=c, B C=a_{1}, C A=b_{1}$ and $A B=c_{1}$. We will compute the length $m$ of the median $D M$. Let $N$ be the midpoint of edge $B C, D N=p$ and $A N=q$. Then $D M^{2}+$ $+M N^{2}-2 D M \cdot M N \cos D M N=D N^{2} \quad$ and $D M^{2}+A M^{2}-$ $-2 D M \cdot A M \cos D M A=A D^{2}$, which means,

$$
m^{2}+\frac{q^{2}}{9}-\frac{2 m q \cos \varphi}{3}=p^{2} \quad \text { and } \quad m^{2}+\frac{4 q^{2}}{9}+\frac{4 m q \cos \varphi}{3}=a^{2}
$$

Multiplying the first equation by 2 and adding it to the second equation, we get $3 m^{2}=a^{2}+2 p^{2}-2 q^{2} / 3$. Since $p^{2}=$ $=\left(2 b^{2}+2 c^{2}-a_{1}^{2}\right) / 4$ and $\quad q^{2}=\left(2 b_{1}^{2}+2 c_{1}^{2}-a_{1}^{2}\right) / 4$, then $9 m^{2}=$ $=3\left(a^{2}+b^{2}+c^{2}\right)-a_{1}^{2}-b_{1}^{2}-c_{1}^{2}$.