# midpoint theorem connections to midpoint formula

Is there an elegant way to prove that the midpoint is $\left(\frac{x_1+y_1}{2},\frac{x_2+y_2}{2}\right)$ using non-right angled similar triangles? I know that it's really easy to just by assuming this is true and proving by showing the distance formula gives two equal sides. Or is there some other way I can relate the midpoint formula to the midpoint theorem (the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side). • The midpoint of what, and what are $x_i,y_i$ ? – Yves Daoust Apr 22 '17 at 8:52
• i think $$x_1,y_1$$ and $$x_2,y_2$$ are coordinates of two points – Dr. Sonnhard Graubner Apr 22 '17 at 8:58
• How do you want to use triangles to prove the above if you only have two points? – DonAntonio Apr 22 '17 at 9:00

The formula for the midpoint of $A(x_1,y_1)$ and $B(x_2,y_2)$ is $$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$
$$\overrightarrow{OM}=\overrightarrow{OA}+\frac 12\overrightarrow{AB}=\overrightarrow{OA}+\frac 12\left(\overrightarrow{OB}-\overrightarrow{OA}\right)=\frac 12\left(\overrightarrow{OA}+\overrightarrow{OB}\right)$$