I have "discovered" this property while playing with the circumscribed circle of a triangle in GeoGebra.
Let $ABC$ be a triangle, $A'$, $B'$, $C'$ the intersections of the corresponding angle bisectors and perpendicular bisectors of $ABC$, opposite to $A$, $B$, and $C$ accordingly.
Then the angle bisectors and perpendicular bisectors of $ABC$ divide the angles of the triangle $A'B'C'$ into three parts, two of which, adjacent to the sides of $A'B'C'$, are equal.
Questions:
- Is there a name for the triangle $A'B'C'$?
- How to prove the equality of the angles?
I think the crucial part of your post is to note the fact that $A'$, $B'$, and $C'$ lie on the circumcircle of the triangle, which is not very hard to be proved (just note that if, for instance, $C'$ is the intersection of the perpendicular bisector of $AB$ and the circumcircle, then $AC'=C'B$). Having this fact, some angle chasing is enough for the second part of your question. For example, according to your image,
$$\angle C'A'O=\angle BA'O -\angle BA'C'=(90^{\circ}-\frac{\angle A}{2})-(\frac{\angle C}{2})=\frac{\angle B}{2}.$$
On the other hand,
$$\angle AA'B'=\angle ABB'=\frac{\angle B}{2}.$$