∠ A bisector intersects the circumscribed circle of △ABC at the midpoint of arc AB;

On the other hand, BC- perpendicular bisector intersection △ABC circumscribes the midpoint of arc AB;

So d is the midpoint of arc AB.

A, B, C, D*** cycles, in the same way, you can get six * * * cycles.

Method 2:

Because BD = DC；;

∠BAD=∠CAD

Therefore, BD/sin∠BAD=DC/sin∠CAD.

Therefore, the radius of the circumscribed circle of △ABD and △ADC is equal, and it is set to R. 。

Then sin∠ABD=AD/(2R)=sin∠ACD.

Because ∠ABD is not equal to ∠ACD, otherwise it is deduced that △ABC is an isosceles triangle.

So ∠ABD+∠ACD= 180, a, b, c, D*** cycles.