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