Solid Geometry – Problem 11: Prove that

In a regular tetrahedron $SXYZW$ (all edges equal), let $A$ and $B$ be the midpoints of edges $SX$ and $SY$, respectively. Let $C$ and $D$ be the midpoints of edges $WX$ and $WY$, respectively.

(1) Prove that $AB\parallel CD$.

(2) Find the angle between segments $AB$ and $CD$.

Use vectors in $\mathbb R^3$. Choose an origin $O$ and denote the position vectors of $S,X,Y,W$ by $\vec s,\vec x,\vec y,\vec w$. Because $A,B,C,D$ are midpoints, $$\vec a=\frac{\vec s+\vec x}{2},\quad \vec b=\frac{\vec s+\vec y}{2},\quad \vec c=\frac{\vec w+\vec x}{2},\quad \vec d=\frac{\vec w+\vec y}{2}.$$ Then $$\overrightarrow{AB}=\vec b-\vec a=\frac{\vec y-\vec x}{2},$$ and $$\overrightarrow{CD}=\vec d-\vec c=\frac{\vec y-\vec x}{2}.$$ So $\overrightarrow{AB}=\overrightarrow{CD}$, which implies $$AB\parallel CD.$$ Therefore the angle between $AB$ and $CD$ is $$0^\circ.$$

$AB\parallel CD$, so the angle between $AB$ and $CD$ is $0^\circ$.