Rectangle ABCD has points E and F on sides AB and CD, respectively. If AE = AB/3 and CF = CD/4 and segments DE and BF intersect diagonal AC at G and H, respectively, what is the ratio AG:GH:HC? Express your answer in the form a:b:c, where a, b and c are relatively prime positive integers.
+128460 
Triangles AGE and CGD are similar
Since AE = (1/3)AB
Then AE = (1/3) CD
So AG = (1/3) CG
CG = GH + HC
And triangles CHF and AHB are similar
Then CF = (1/4)CD = ( 1/4) AB
So HC = (1/4) HA
HC = (1/4) ( GH + AG)
AG = (1/3)CG
AG = (1/3) (GH + HC)
3AG = GH + HC
HC = 3AG - GH
Therefore
HC = HC so
3AG - GH = (1/4)(GH +AG)
3AG - GH = (1/4)GH + (1/4)AG
(3 - 1/4)AG = ( 1 + 1/4) GH
(11/4)AG = (5/4)GH
11AG = 5GH
AG = (5/11)GH
And HC = (1/4)(GH + AG) = (1/4) [GH + (5/11) GH ] = (1/4)(16/11)GH = (16/44)GH = (4/11)GH
So
AG : GH : HC =
(5/11)GH : GH : (4/11) GH =
5GH : 11 GH : 4GH =
5 : 11 : 4
