I won't repeat the part of the first question that needs to be proved, and go directly to the second question. If you don't understand anything, ask me again.

Question 2: connect BD, DF and CF, and let the radius of the circle be R.

∠ ADB =? ∠ACB？ =? 60 and BE⊥AD

Then ∠DBE= 30, then ∠KBD? =? 60. Triangle KBD is an equilateral triangle.

BE is the height on the cardinal number KD, so it is also the center line, KE=DE.

Right triangle DEF, ∠ADF? =? ∠ABF？ =? 45, so it is an isosceles right triangle with DE=EF=KE.

It can be concluded that DFK is also an isosceles right triangle.

Launch ∠KFE =? 45

Launch ∠KFH =? 15 ? =? ∠CBF

∠KHF？ =? ∠CFB？ =? 90

The triangle KHF is similar to the triangle CFB.

KF/BC=FH/BF=sin30 = 1/2

So KF = 0.5 * BC = R.

OF is the radius of a circle, so of = R.

So KF =?

The answer to the third question