Prove that $AC \perp CE$.

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, $\angle 1 = \angle 4= \angle 5 = \angle 6$.

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

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 $AC \perp CE$.