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Ninth grade mathematics rectangle
Let AB = 2K(K & gt;; 0), then AD=BC=3K, be =1.5k.

∵ABCD is a rectangle, ∴AD∥BC, ∴∠DAF=∠AEB,

∵DF⊥AE,∴∠AFD=∠B=90,

∴δadf∽δeab,

∴AF/DF=BE/AB= 1.5K/2K=3/4,

And af 2+df 2 = ad 2,

∴AF^2+ 16/9AF^2=9K^2,

AF= 1.8K,

∵AE=√(AB^2+BE^2)=2.5K,

∴EF=0.7K,

Extend DF to CB in h,

∵ad∥bc,∴δadf∽δehf,

∴eh/ad=ef/af==0.7k/ 1.8k=7/ 18,

∴EH=7/6K,∴CH=7/6K+3/2K=8/3K,

∴AG:CG=AD:CH=3K:8/3K=9:8。