# 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?

• In your notation, you're saying that, for example, $\vec{AB} = \vec{A} + \vec{B}$. Are you sure that's right? Commented Nov 28, 2021 at 20:05

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.

In the Cartesian plane, we have vertices $$\vec{A}, \vec{B}, \vec{C}$$ as $$2$$-vectors (i.e. vectors having $$2$$ components).

Define $$\vec{u} = \vec{B} - \vec{A}$$ and $$\vec{v} = \vec{C } - \vec{A}$$

Thus $$\vec{B} = \vec{A} + \vec{u}$$ and $$\vec{C} = \vec{A} + \vec{v}$$

The midpoint of $$AB$$ is

$$\vec{M_1} = \dfrac{1}{2} (\vec{A} + \vec{B} ) = \dfrac{1}{2} ( 2 \vec{A} + \vec{u} ) = \vec{A} + \vec{1}{2} \vec{u}$$

The midpoint of $$BC$$ is

$$\vec{M_2} = \dfrac{1}{2} (\vec{B} + \vec{C} ) = \dfrac{1}{2} ( 2 \vec{A} + \vec{u} + \vec{v} ) = \vec{A} + \dfrac{1}{2} ( \vec{u} + \vec{v} )$$

Hence the vector extending from $$\vec{M_1}$$ to $$\vec{M_2}$$ is given by

$$\vec{W} = \vec{M_2} - \vec{M_1} = \dfrac{1}{2} \vec{v}$$

Since $$\vec{v} =\vec{AC} = \vec{C} - \vec{A}$$, then

$$\vec{W}$$ is parallel to $$\vec{AC}$$ and is half its length.