So, given a triangle with points A, B, C.


ecuations of two medians are.

$x-2y+1=0$ and $y-1=0$

I found that all medians intersect at point $(1,1)$ , so i can find equation for third median that goes through $A$ point.

But how to find three equations of each side of the ABC triangle?


I'll give you an idea on how to construct the lines. Suppose we have one of those lines. Then we know it intersects one median at one of the vertices of the triangle, let's just say it will be $B$, and the other median at the midpoint between $A$ and $B$. By the way, let's call the intersection of medians $M$.

So after looking at it a bit you can see that if you draw a parallel to the second median at $A$ and intersect it with the second median, you'll get some point $P$. If you draw a line through $P$ orthogonal to the second median, you'll get the point $B$. Why? Because $AMBP$ is a rectangle (green in the picture), so the first median, going through $M$ and $P$, goes through the midpoint $M_1$ of $AB$.

You can proceed to find $C$ by mimicking the first approach, however now you won't get a rectangle but a parallelogram.

enter image description here

  • $\begingroup$ in what software you have drawn this? :) $\endgroup$ – Daniel Sep 27 '15 at 14:53
  • $\begingroup$ I used GeoGebra :) $\endgroup$ – gus Sep 27 '15 at 15:32

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