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
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).
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
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 © 1996-2018 Alexander Bogomolny