This theorem can be proven by placing the triangle on a coordinate plane. For simplicity, vertex $B$ will be placed at the origin and vertex $C$ on the $x\text{-}$axis.

Since $B$ lies on the origin, its coordinates are $(0,0).$ Point $C$ is on the $x\text{-}$axis, meaning its $y\text{-}$coordinate is $0.$ The remaining coordinates are unknown and can be named $a,$ $b,$ and $c.$ If $\overline{DE}$ is the midsegment from $\overline{BA}$ to $\overline{CA},$ then by the definition of a midpoint, $D$ and $E$ are the midpoints of $\overline{BA}$ and $\overline{CA},$ respectively.

To prove this theorem, it must be proven that $\overline{DE}$ is parallel to $\overline{BC}$ and that $DE$ is half of $BC.$

$\overline{BC}\parallel \overline{DE}$

If the slopes of these two segments are equal, then they are parallel. The $y\text{-}$coordinate of both $B(0,0)$ and $C(a,0)$ is $0.$ Therefore, $\overline{BC}$ is a horizontal segment. Next, the coordinates of $D$ and $E$ will be found using the Midpoint Formula.

$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
Segment	Endpoints	Substitute	Simplify
$\overline{BA}$	$B(0,0)$ and $A(b,c)$	$D\left(\frac{0+b}{2},\frac{0+c}{2}\right)$	$D\left(\frac{b}{2},\frac{c}{2}\right)$
$\overline{CA}$	$C(a,0)$ and $A(b,c)$	$E\left(\frac{a+b}{2},\frac{0+c}{2}\right)$	$E\left(\frac{a+b}{2},\frac{c}{2}\right)$

The $y\text{-}$coordinate of both $D\left(\frac{b}{2},\frac{c}{2}\right)$ and $E\left(\frac{a+b}{2},\frac{c}{2}\right)$ is $\frac{c}{2}.$ Therefore, $\overline{DE}$ is also a horizontal segment. Since all horizontal segments are parallel, it can be said that $\overline{BC}$ and $\overline{DE}$ are parallel.

$DE=\frac{1}{2}BC$

Since both $\overline{BC}$ and $\overline{DE}$ are horizontal, their lengths are given by the difference of the $x\text{-}$coordinates of their endpoints.

Segment	Endpoints	Length	Simplify
$\overline{BC}$	$B(0,0)$ and $C(a,0)$	$BC=a-0$	$BC=a$
$\overline{DE}$	$D\left(\frac{b}{2},\frac{c}{2}\right)$ and $E\left(\frac{a+b}{2},\frac{c}{2}\right)$	$DE=\frac{a+b}{2}-\frac{b}{2}$	$DE=\frac{1}{2}a$

Since $\frac{1}{2}a$ is half of $a,$ it can be stated that the midsegment $\overline{DE}$ is half the length of $\overline{BC}.$ Therefore, a midsegment of two sides of a triangle is parallel to the third side of the triangle and half its length.

This proof will be developed based on the given diagram, but it is valid for any triangle. To prove this theorem, it must be proven that $\overline{DE}$ is parallel to $\overline{BC}$ and that $DE$ is equal to half of $BC.$ Each statement will be proven one at a time.

$\overline{BC}\parallel \overline{DE}$

This part can be proven by using rigid motions. First, translate $△ADE$ along $\overline{DB}$ so that $D$ is mapped onto $B.$ Since $D$ is the midpoint of $\overline{AB},$ $A$ is mapped onto $D.$ Next, it must be proven that the image of $E$ — which is marked as $E^{\prime }$ — lies on $\overline{BC}.$ This proof will be done using indirect reasoning. In this method, it is temporarily assumed that the negation of the statement is true. It has been proven that $E^{\prime }$ lies on $\overline{BC}.$ Because translations preserve angles, $\angle ADE$ is congruent to $\angle DB{E}^{\prime }.$

By the Converse Corresponding Angles Theorem, $\overline{BC}$ is parallel to $\overline{DE}.$

$DE=\frac{1}{2}BC$

It has been previously obtained that $\overline{BC}$ and $\overline{DE}$ are parallel. Now, another rigid motion to $△BD{E}^{\prime }$ will be applied. By the Segment Addition Postulate, the length of $\overline{BC}$ can be calculated by adding the lengths of smaller segments. Because corresponding sides of congruent triangles are congruent, $\overline{DE}$ is congruent to $\overline{B{E}^{\prime }}$ and $\overline{DE}$ is congruent to $\overline{C{E}^{\prime }}.$ Therefore, $DE=B{E}^{\prime }$ and $DE=C{E}^{\prime }.$ By the Substitution Property of Equality, $BC$ can be expressed in terms of $DE.$ Finally, by the Division Property of Equality, the second statement of the theorem is obtained.