Prove that $AC \perp CE$. $ ($MB < MC$). $MO$ intersects $AD$ at $E$. Prove that $AC \perp CE$. I am trying to prove that $\widehat{MAB} = \widehat{ECD}$ because $$\widehat{MAB} = \widehat{ACB} = 90^\circ - \widehat{DCA} = \widehat{ECD}$$

Let $F$ be such that $AF$ is a diameter of the circle. Using Pascal theorem for (degenerate) hexagon $BDAAFC$ we obtain that $M$, $O$, and the intersection of $FC$ with $AD$ are collinear. In other words, $MO$, $FC$, and $AD$ are concurrent, i.e. $E$, $C$, and $F$ are collinear. Since $AF$ is a diameter, it follows that $AC \perp CE$.