Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third and half as long

Question

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:

Answer 1

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.

Answer 2

\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}