six planes containing one edge and bisecting the opposite edge of a tetrahedron Show that the six planes containing one edge and bisecting the opposite edge of a tetrahedron meet in a point.
I have no idea to solve it using vector algebra.
 A: Let $O$, $A$, $B$, $C$ be the four vertices of the tetrahedron, and define:
$$
\vec a=A-O,\quad\vec b=B-O,\quad\vec c=C-O.
$$
Let now $M=(B+C)/2=O+(\vec b+\vec c)/2$ be the midpoint of edge $BC$. The plane passing through edge $OA$ and point $M$ is parallel to vectors $\vec a$ and 
$(\vec b+\vec c)/2$, thus any point $P$ on it satisfies
$$
P-O=\vec a s +{\vec b+\vec c\over2}t,
$$
where $s$ and $t$ are two real parameters.
With a similar reasoning we can find the equation for the “dual” plane, passing through edge $BC$ and the midpoint of $OA$:
$$
P-O=\left(\vec b-{\vec a\over2}\right) s +\left(\vec c-{\vec a\over2}\right) t+{\vec a\over2}.
$$
To get convinced that this equation is correct, just observe that $B$ belongs to the plane (for $s=1$ and $t=0$), $C$ belongs to the plane (for $s=0$ and $t=1$), and $(A+O)/2$ belongs to the plane (for $s=t=0$).
If $G=(O+A+B+C)/4=O+(\vec a+\vec b+\vec c)/4$ is the centroid of the tetrahedron, you can easily check that $G$ lies on both planes: you get $P-O=(\vec a+\vec b+\vec c)/4$ by letting $s=1/4$, $t=1/2$ in the first equation, and $s=t=1/4$ in the second equation.
The equations of the other four planes can be obtained from the two equations above, by making the substitutions: $\vec a\to \vec b$, $\vec b\to \vec c$, $\vec c\to \vec a$ for third and fourth plane, and the substitutions: $\vec a\to \vec c$, $\vec b\to \vec a$, $\vec c\to \vec b$ for fifth and sixth plane. 
As the expression for $G$ is invariant under such substitutions, we are guaranteed that $G$ belongs to all six planes, thus proving the given statement.
A: Using barycentric coordinates. 
The points in the plane described by three non-collinear points $P$, $Q$, $R$ are of the form $\alpha P + \beta Q + \gamma R$, where $\alpha + \beta + \gamma = 1$.
Take the plane through the points $A$, $B$, and the middle of the side $CD$, that is , $\frac{1}{2}C + \frac{1}{2}D$. We have 
$$\frac{1}{4}A + \frac{1}{4}B + \frac{1}{2}\cdot ( \frac{1}{2} C + \frac{1}{2} D) = \frac{1}{4}A + \frac{1}{4}B + \frac{1}{4}C + \frac{1}{4} D$$ so that plane contains the center of mass of the tetrahedron. 
This also suggests a geometric approach: take the middle of the segment with ends the middle of $AB$ and $CD$. It will lie in the plane, and it is the same point for all the planes.  We can see this geometrically: Let $M$ be the middle of $AB$, $N$ the middle of $CD$, $P$ the middle of  $AD$, $Q$ the middle of $BC$. Then $MP$ is parallel to $BD$ and $\frac{1}{2}$ of it; same for $QN$. Therefore, $MPNQ$ is a parallelogram, and so its diagonals $MN$ and $PQ$ intersect in the middle. 
