ABC is a triangle. G is the centroid. The line parallel to BC through G meets AB at B' and AC at C'. Let A'' be the midpoint of BC, C'' the intersection of B'C and BG, and B'' the intersection of C'B and CG. Prove that A''B''C'' is similar to ABC.
Solution
Let M be the midpoint of AB and N the midpoint of AC. Let A''M meet BG at X. Then X must be the midpoint of A''M (an expansion by a factor 2 center B takes A''M to CA and X to N). Also BX/BN = 1/2 and BG/BN = 2/3, so XG = BX/3. Let the ray CX meet AB at Z. Then ZX = CX/3. (There must be a neat geometric argument for this, but if we take vectors origin B, then BX = BN/2 = BA/4 + BC/4, so BZ = BA/3 and so XZ = 1/3 (BA/4 - 3BC/4) = CX/3.) So now triangles BXC and ZXG are similar, so ZG is parallel to BC, so Z is B' and X is C''. But A''X is parallel to AC and 1/4 its length, so A''C'' is parallel to AC and 1/4 its length. Similarly A''B'' is parallel to AB and 1/4 its length. Hence A''B''C'' is similar to ABC.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002