Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third and half as long The task is to prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long. (Or in vector notation 
PQ = AB / 2). It should be proved using some vector algebra but I am not sure how to go about doing it. 
A (crude) visualization:
 A: \begin{equation}
\overrightarrow{PQ}=\overrightarrow{CQ}-\overrightarrow{CP}=\dfrac{1}{2}\overrightarrow{CB}-\dfrac{1}{2}\overrightarrow{CA}=\dfrac{1}{2}\left( \overrightarrow{CB}-\overrightarrow{CA}\right)=\dfrac{1}{2}\overrightarrow{AB}  
\end{equation}
A: With only basic geometry:
If you've already studied similarity of triangles it is pretty easy: comparing triangles $\;\Delta CAB\,,\,\,\Delta CPQ\;$ :
$$\frac12=\frac{CP}{CA}=\frac{CQ}{CB}\;,\;\;\text{and the angle}\;\; \angle C\;\;\text{is common to both triangles}$$
By similarity theorem , $\;\Delta CAB\sim\Delta CPQ\;$ , and thus
$$\frac{PQ}{AB}=\frac12\iff 2PQ=AB$$ 
That $\;PQ||AB\;$ follows from the fact that similar triangles have the same angles, and thus $\;\angle CAB=\angle CPQ\;$ .
With vectors:
Put $\;u:=\vec{CA}\;,\;\;v:=\vec{CB}\;$ , then we get:
$$\vec{CP}=\frac12u\;,\;\;\vec{CQ}=\frac12CB\;,\;\;\vec{AB}=-u+v=-(v-u)$$
so
$$\vec{PQ}=-\frac12+\frac12b=-\frac12(u-v)=\frac12\vec{AB}$$
and we're done as the last line both proves the middle segment is parallel to $\;AB\;$ and its length is half that of the latter.
