G is the cenetroid of triangle ABC and D and E are the midpoints of line AB and line AC, repectively. Line AG and line DE intersect at M. Find GM/GA.
+128460 Draw FH through G such that FH is parallel to DE.....and let F lie on AB and H lie on AC
Then triangles DEA , FHA and BCA are similar
And since G is a median then AG is 2/3 of the distance from A to BC
So...AF is 2/3 of AB
And AH is 2/3 of AC
So DF = (2/3)AB - (1/2)AB = (1/6)AB
So DF / AF = [ (1/6)AB] / [(2/3)AB] = (1/6) / (2/3) = 1/4
And because triangles DEA and FHA are similar....then DF /AF = GM / GA = 1/4
