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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).

example of this if a line were joined through $m_1$ and $m_2$

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  • $\begingroup$ The midpoint of what, and what are $x_i,y_i$ ? $\endgroup$ – Yves Daoust Apr 22 '17 at 8:52
  • $\begingroup$ i think $$x_1,y_1$$ and $$x_2,y_2$$ are coordinates of two points $\endgroup$ – Dr. Sonnhard Graubner Apr 22 '17 at 8:58
  • $\begingroup$ How do you want to use triangles to prove the above if you only have two points? $\endgroup$ – DonAntonio Apr 22 '17 at 9:00
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The formula for the midpoint of $A(x_1,y_1)$ and $B(x_2,y_2)$ is $$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$

The simplest proof of this uses vectors.

$$\overrightarrow{OM}=\overrightarrow{OA}+\frac 12\overrightarrow{AB}=\overrightarrow{OA}+\frac 12\left(\overrightarrow{OB}-\overrightarrow{OA}\right)=\frac 12\left(\overrightarrow{OA}+\overrightarrow{OB}\right)$$

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To add to David Quinn:

Consider a line in your diagram passing through the midpoint of AB running parallel to AC. Where does it intersect BC?

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