The Medians

Dear Alex,

Here is yet another approach to the concurrence of the medians of a triangle (or at least of the demonstration that the medians cut each other in the ratio of 1:2). It is a corrected version of an argument which my son, Joshua, brought home from his ninth grade geometry teacher, Charles Worrall of Horace Mann School.

Your page mentions that the six smaller triangles into which the medians divide the triangle are equal in area. Of course, that statement is "meaningful" only if it has already been established that the medians are concurrent. To avoid circularity, consider the attached figure: in triangle ABC, AE and BF are medians, which intersect at point G; GD is the median of triangle ABG; do NOT assume that GD and CG are collinear. The subsidiary triangles are numbered clockwise in succession -- let each number also represent the AREA of its triangle.

Applying Euclid I.38 ("Triangles which are on equal bases and in the same parallels equal one another.") repeatedly, observe 1 = 2, 3 = 4, 5 = 6. Likewise, 4 + 3 + 2 = 5 + 6 + 1. Subtracting 1 = 2 gives 4 + 3 = 5 + 6; substituting 3 = 4, 5 = 6, 4 + 4 = 5 + 5, or 4 = 5. Repeating the argument starting from 6 + 1 + 2 = 5 + 4 + 3 yields 2 = 3; thus all six triangles are equal in area.

Now apply Euclid VI.1 ("Triangles and parallelograms which are under the same height are to one another as their bases."): Comparing triangle 2 with the union of 3 and 4, it follows that BG = 2GF; similarly AG = 2GE.

Therefore etc. One more instance of a proof by comparing areas as an alternative to a proof by similarity (a situation similar to that encountered with the Pythagorean Theorem).

Best Regards,
        Scott Brodie

Remark

Let me remark on what might be a natural ending for Scott's argument. "Therefore etc." assumes familiarity with the proofs on the main Medians page. The argument is similar to that in the first section.

Scott shows that the point of intersection of the two medians AE and BF devides each of them in the ratio of 2:1. This ratio determines a unique point on both of AE and BF. Assume now we selected a different pair of medians, to start with. The argument will go through for this pair as well. One of the original medians (either AE or BF) will be included into the new pair. Therefore, its "2:1" point is shared by all three medians.

This is a more explicit statement of the argument "by symmetry" employed in the first of the elementary proofs, see above. Transitivity of equality pops up naturally in this context.

Copyright © 1996-2008 Alexander Bogomolny