My pure geometric solution:

Let ∠A=3α, ∠B=3β, ∠C=3γ of △ABC.

Two bisectors adjacent to BC intersect at x,

The other two bisectors of ∠B and ∠C intersect at S,

Then x is the heart of △SBC, so XS shares ∠BSC equally.

∠ sxz =∠ sxy = 30 on both sides of sx,

Z and y are on BS and CS respectively, then △ sxz △ sxy,

So XZ=XY, and ∠ zxy = 60, so △XYZ is an equilateral triangle.

It is proved that AZ and AY are divided into three parts ∠A, BX'=BX and CX "= CX are intercepted on BA and CA respectively.

Then △ bzx' △ bzx, so ZX'=ZX=ZY. Similarly, there is YX"=ZY, so X'Z=ZY=YX ".

∠X'ZY=360 -2∠BZX-60

∠X'ZY=360 -2( 1/2∠S+30 )-60

∠X'ZY=240 -∠S

∠X'ZY=240 -( 180 -2β-2γ)

∠X'ZY=60 +2(β+γ)

∠X'ZY=60 +2(60 -α)= 180 -2α

It can also be proved that ∠ zyx "= 180-2α.

As the circumscribed circle o of △X'ZY, we can know from symmetry that x "is on the circumscribed circle o..

Yi Zheng's central angle ∠X'OZ=∠ZOY=YOX"=2α. Therefore, ∠X'OX"=6α.

And because ∠A=3α, point A is also on circle 0.

The bisector of AZ and AY∠A is obtained by using the chord line X'Z=ZY=YZ ",thus proving the proposition.

The second topic should be the butterfly theorem.

I use a simple elementary math to prove it.

Please ask questions if you don't understand, and let's discuss them together. I hope I can help you!