∴AB=AD,∠BAD=90？ .

∵DE⊥AG，∴∠AED=90？ . And ∵BF∨de, ∴∠AFB=∠DEA=90? ,

∠∠DAE+∠BAF = 90？ ，∠ABF+∠BAF=90？ ,∴∠DAE=∠ABF.

In δ ABF and δ DAE, ∠ABF=∠DAE, ∠BFA=∠AED, AB=DA,

∴δabf≌δdae,∴ae=bf,∴ef=af-ae=af-bf

16, the length relationship between BO and OD is BO = 2OD for the following reasons:

Take the midpoint m of BO aND the midpoint n of CO to connect ED, EM, MN, nd,

∴MN is the center line of δδBOC, ED is the center line of δδABC,

∴MN∥BC,MN=？ BC，ED∨BC，ED=？ BC,

∴MN∥ED, and MN=ED, ∴ quadrilateral EMND is a parallelogram, ∴MO=OD.

∵BM=MO,∴BO=2OD。

The midline on the BC side must pass through point O for the following reasons:

Let the median line of BC side intersect BD at point O',

We know that bo ′ = 2o ′ d.

BO = 2OD，

∴ point o coincides with point o', that is, the center line of BC side must pass through point o.

No points? ! ! !