Prove that $AC \perp CE$. 

$MA$, $MBC$ and $BD$ is respectively a tangent, a secant and a diamter of $(O)$ ($MB < MC$). $MO$ intersects $AD$ at $E$. Prove that $AC \perp CE$.

I am trying to prove that $\widehat{MAB} = \widehat{ECD}$ because
$$\widehat{MAB} = \widehat{ACB} = 90^\circ - \widehat{DCA} = \widehat{ECD}$$
 A: Let CE cut circle (O) at Z. Join AZ. The tricky part is that at this stage AOZ is not a straight line yet but AZ is.
$\angle ADZ = \angle 5 + \alpha’ + \beta’ = \angle 7 + \alpha + \beta = 90^0$. This means AZ is a diameter of the circle (O) and hence AOE is a straight line.

Extend BD to some point X such that $XE \bot DE$. Extend XE to meet MC at Y.
It should be clear that CDEY is cyclic with DY as diameter.
Since ZD //YX, $\angle 1 = \angle 2 = \angle 3$. This means YD is tangent to the circle (O) at D. 
The above further implies BDX is tangent to the circle (K) at D. Then, $\angle 1 = \angle 4= \angle 5 = \angle 6$. 

Remark: The proof is complete after the first 2 paragraphs. The rest is just additional work showing how to get $\angle 1 = \angle 6$.
A: 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 $AC \perp CE$.
A: Hint


Prove that the triangles $\triangle ABC$ and $\triangle CED$ are similar or equivalently that $\angle ECD=\angle ACB=\alpha$.

Thus $$\angle ECA=\alpha+\angle DCA=\alpha+\angle DBA=\alpha+\angle BAO=\color{blue}{\angle MAO=90°\implies AC\perp CE}$$
