1, (1) connects AE, ∵E and f are the midpoint of BC and AC respectively, ∴EF∥AB, EF= 1/2AB.

∫AD = 1/2ab, ∴AD=EF, ∴ quadrilateral AEFD is a parallelogram (ad and EF are parallel and equal).

∴DF=AE。 (If ABC has no other conditions and AE has nothing to do with BE, please check whether the conclusion is wrong).

(2) (There is no quantitative relationship between AG and DG under no other conditions), but it can be proved that DG=FG.

If f is FH∑BC and AB is H, then H is the midpoint of A, ∴AH=BH=AD,

And ∵ag∨BC, ∴AG∥FH, ∴AG is the midline of Δ δδdah, ∴DG=FG.

2. extend the intersection of CE and AB to f, ∫∠EAF =∠EAC, ∠ ADC = ∠ ADF = 90, and AD=AD.

∴δadf≌δadc, ∴CE=EF, AF=AC, ∴E is the midpoint of CF, BF=AB-AC.

∫n is the midpoint of BC, ∴EN is the centerline of δδCFB, ∴ en =1/2bf =1/2 (AB-AC).

3.ac = ad, AE⊥CD, ∴CE=DE, and F is the midpoint of BC.

∴EF is the center line of∴ BD = 2ef. δ δ CDB.