I studied ∠ Cao = ∠ Bao = & gtOA is the bisector of ∠BAC.

Circles o and AB are tangent to point b = & gtOB⊥AB

OC=OB=r

The above three conditions can be known from the theorem of angular bisector (the distance from any point on the angular bisector to both sides of this angle is equal): oc ⊥ AC = > AC is tangent to circle O.

2. According to the circle theorem: OA⊥ and bisecting BC, according to the problem: OA=6 r=2, so AC=AB=4√2 is easy to prove:

△COD∽△AOC, and the relationship between the two sides is: CD=4√2/3, BC=8√2/3.

Let BE=x Pythagorean Theorem (in △BCE and △ACE) both seek CE, so there are two equations to seek x=8√2/9.

In △BCE, on the side of △CDF∽△CEB, CE=32/9, CF=2.

So EF= 14/9.

Or connect BF, and see if it can be concluded that the quadrangle OCFB is a diamond, and CF=2 directly. (But I always feel that it is impossible, and there are fewer conditions. )