IMO Team Selection Test 2

We first prove that $△CMB\sim △DBE$. Since $D$ and $E$ lie on opposite sides of $MB$, it holds that $\angle DBE=\angle DBM+\angle MBE=\angle DBM+\angle MDB=\angle CMB$ because of the given similarity and the exterior angle theorem. Moreover, it holds that $$\frac{|DB|}{|BE|}=\frac{|MD|}{|MB|}=\frac{|CM|}{|MB|}$$ because of the similarity defining $E$ and the fact that $M$ is the midpoint of $CD$. It now follows that $△CMB\sim △DBE$ (sas). In particular, it follows that $\angle BED=\angle MBC$. Therefore $\angle ABD=\angle MBC$ if and only if $\angle ABD=\angle BED$. By the inscribed angle theorem (tangent case), this holds if and only if $AB$ is tangent to the circumcircle of $△BDE$. The circle through $B$ and $D$ tangent to $AB$ is unique, and has as centre the intersection of the perpendicular bisector of $BD$ and the line through $B$ perpendicular to $AB$. So $AB$ is tangent to the circumcircle of $△BDE$ if and only if $E$ lies on $\Omega$. $\square$