∴∠BDC=∠BEC, that's ⊿BEC∽⊿BDC.

∴BD/BG=BE/BC，

BG*BE= BD*BC

(2)∫d is ∴BC=2BD. The midpoint of BC.

∵⊿ABC is an isosceles right triangle, ∴AB=√2BD.

That is BG * be = 2bd 2 = (√ 2bd) 2 = ba 2.

∴BG/BA=BA/BE

In ⊿BAE and ⊿BGA, Abe = abg.

∴⊿BAE∽⊿BGA, that is ∠ BAE = ∠ BGA = 90.

∴AG vertical BE

(3) Connect DE, where E is the midpoint of AC and D is the midpoint of BC. ∴DE//BA. Because BA⊥AC, DE⊥AC sets AB=2a AE=a as the extension line of the intersection of CH⊥BE at H.

∠∠AEG =∠CEH, ∠ Age =∠∠ Cut, AE=EC.

∴△AEG≌△CEH(AAS)∴CH=AG ∠GAE=∠HCE

∵∠BAE is a right angle ∴ BE =∴ 5A ∴ AE = AB * AE/BE = (2/∴ 5) A ∴ CH = (2/∴ 5) A.

∵ag⊥be,∠fge=45∴∠agf=45=∠ecb∵∠dfe=∠gae+∠agf=∠hce+∠ecb； ∴∠DFE=∠BCH

And CH⊥BE ∴△DEF∽△BHC.

∴ef:df=ch:bc=(2/√5)a:2√2a= 1:√ 10=√ 10/ 10