Triangle Center Midpoint Proof

I am working through problem 45, Stewart's calculus 6e.

Use vectors to prove that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

Ultimately, I am trying show the midpoint vector (line between the triangle's two midpointsenter image description here) is parallel and half the size. I have included a picture labeling everything.

The three vertices of an arbitrary triangle as vectors $\vec{A} = < a_1,a_2>$, $\vec{B} = < b_1,b_2>$, $\vec{C} = < c_1,c_2>$.

Then connect the vertices to make the lengths of the triangle: $\vec{AB} = < a_1 + b_1,a_2 + b_2>$, $\vec{BC} = < b_1 + c_1 ,b_2 + c_2>$, $\vec{CA} = < c_1 + a_1 ,c_2 + a2>$.

Next we want to define the midpoints of $AB_m = < \dfrac{a_1 + b_1}{2} , \dfrac{a_2 + b_2}{2}>$ and $BC_m = < \dfrac{b_1 + c_1}{2} , \dfrac{b_2 + c_2}{2}>$

We want to show that $\vec{AB_mBC_m}$ is half the size of $\vec{AC}$.

$\vec{AB_mBC_m} = < \dfrac{a_1 + b_1}{2} + \dfrac{b_1 + c_1}{2}, \dfrac{a_2 + b_2}{2} + \dfrac{b_2 + c_2}{2} >$

We factor the vector equation by $\dfrac{1}{2}$ :

$\vec{AB_mBC_m} = \dfrac{1}{2} < a_1 + b_1 + b_1 + c_1, a_2 + b_2 + b_2 + c_2 >$

$\vec{AB_mBC_m} = \dfrac{1}{2} < a_1 + 2b_1 + c_1, a_2 + 2b_2 + c_2 >$

But this is not equal to $\dfrac{1}{2}$ of $\vec{CA}$. Did I write something incorrectly or am I missing something?

If the triangle has vertices at $\vec a$, $\vec b$, and $\vec c$, then two midpoints could be $p_1=\frac{\vec a+\vec b}2$ and $p_2=\frac{\vec b+\vec c}2$. Hence, the displacement vector from $\vec p_1$ to $\vec p_2$ is $$ \vec p_2-\vec p_1=\frac{\vec c-\vec a}2,$$ which satisfies the statement of the problem.

I did not check though what went wrong with your attempted answer though.