# The center of circle passing through the midpoints of sides of isosceles triangle ABC

The center of the circle passing through the midpoints of sides of isosceles triangle $$ABC$$ lies on the circumcircle of triangle $$ABC$$. If largest angle of the triangle is $$x$$ and smallest is $$y$$. find $$x-y$$.

The center of any circle with $$D$$ and $$E$$ on it must pass through the (potentially extended) bisector of $$\angle A$$. For this center to be on the circumcircle of $$\triangle ABC$$, the only possibility is for the center to be $$A$$ itself.
$$AF=AD$$ since they are both radii of the same circle. $$AD=DB$$ since $$D$$ is the midpoint of $$\overline{AB}$$. $$\overline{AF}\perp \overline{BC}$$, since $$\triangle ABC$$ is isosceles. Therefore, since $$AB=2AF$$, $$\angle B=30^\circ$$. That makes $$\angle A=120^\circ$$, so the difference between them is $$90^\circ$$.