(Rigorously) Prove that two medians of a triangle intersect

  • #1
I'm trying to solve exercise 60 from Kiselev's Planimetry.

Homework Statement



"Show that in any triangle, every two medians intersect. Is the same true for every two bisectors? altitudes?"

Homework Equations



There aren't any, unfortunately.

The Attempt at a Solution



The last part of the question is the easiest to answer. The conjecture isn't true for altitudes, as it's possible to have a triangle none of whose altitudes intersects another unless produced (imagine an isosceles obtuse triangle whose vertex has an angle, say, of 120 degrees).

It's intuitively clear that the two medians of a triangle intersect, but I'm having trouble expressing it rigorously.

Here's one approach: one median of a triangle splits the triangle in two, with two vertices and sides on each side of the median. Thus, any other median of the triangle must of necessity intersect the first, because the vertex and the side whose midpoint it joins lie on opposite sides of the first median.

I have two problems with this: the first is that isn't a proof, but perhaps just the beginning of the intuition behind one. The second is that this lacks all rigour.

Here's my attempt to make this idea rigorous:

If I can prove that the median of a triangle lies entirely inside the triangle, then I have a proof, because as the first median splits the interior of the triangle in two, separating both vertices and their opposite sides, forcing the second median to lie in the interior would also force it to intersect the first.

One way I've thought of to make this so is to consider the triangle a pentagon, with the two midpoints in question constituting the additional vertices. The pentagon so formed would be convex, and it is a property of convex polygons that diagonals (or a line segment joining any two points on the boundary, really) lie entirely inside the polygon.

To prove this, I'll have to show that the diagonals of a convex polygon lie entirely in its interior.

Having to do all this to solve a simple elementary geometry problem seems a stretch too far. Is there something I'm missing here?
 

Answers and Replies

  • #2
Simon Bridge
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It does seem that the proof is kinda blowing up as you consider more general cases ... you want to try showing that the situation you have is a special case of something already proven or build from simpler structures.

Maybe:
A line through a vertex, which also intersects the opposite side, can only intersect the triangle at the opposite side (or it would have to pass through another side twice).
Alternatively - any line through a vertex, that does not pass through the opposite side, does not pass through the triangle at all.
That help?

It's certainly the case that it is often easier to rigorously disprove something than to prove it.
 
  • #3
Thanks for the approaches. I'll try them out.

(I suspect it may not be easily possible given what's been presented up to that point in the book, but again, I could just be missing something completely. If I'm still stumped later, I'll write to the author (the translator, actually).)
 

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