In the RT triangle ABC, the angle BAC = 90, AD is perpendicular to BC at point D, O is a point on the side of AC, BO and AD are connected at point F, and OE is perpendicular to OB and BC at point E. ..

(1) Verification: the triangle ABF is similar to the triangle COE；;

(2) When O is the midpoint OF AC edge and AC/AB=2, find the values of of and OE, as shown in Figure 2.

(3) When O is the midpoint of AC edge and AC/AB=n, please write the value of OF/OE directly.

1, proving that:

∠∠BAC = 90 ,∴∠bad=∠bac-∠cad=90-∠CAD，

∵AD⊥BC,∴∠C=90 -∠CAD,∴∠BAD=∠C,①

∵ OE ⊥ OB, ∴∠ BOE = 90, ∴∠ COE =180-90 degrees -∠AOB=90 degrees -∠AOB.

∠ ABO = 90-∠ AOB，∴∠ ABO = ∠ COE ②。

From ① ②, it can be concluded that △ABF and △COE are similar.

2、

Let AB = 1. , then AC=2m, ∫o is the midpoint of AC, ∴ OA = OC = M.

∴OB=, BC =

From the similarity between △ABF and △COE, let BF = X, then OE =,

∴BE=

It is easy to prove that △ABD and △CBA are similar, ∴, ∴ BD =

It is easy to prove that △BDF is similar to △BOE.

Solve,

∴

∴

while

3、

while