Question 99
Given a circle $\omega$ with center $I$. Two Lines Touching $\omega$ at points $A$ and $B$ intersect at point $O$. a point $C$ Is Chosen on the Smaller Arc $AB$ so that $AC \neq CB$. lines $AC$ and $OB$ intersect at point $D$, while lines $BC$ and $OA$ intersect at point $E$. Prove that the Circumcenters of Triangles $ACE$, $BCD$, and $OCI$ Are Collinear. straight lines touching the circle $\omega$ at points $A$ and $B$, intersect at a point $O$. Dot $I$ - Center $\omega$. on a smaller arc $AB$ circles $\omega$ Point selected $C$, different from the middle of the arc. straight $AC$ and $OB$ intersect at a point $D$, and straight $BC$ and $OA$ - at the point $E$. Prove that the centers of the described circles of triangles $ACE$, $BCD$ and $OCI$ lie on one straight line.