It's well known that the intersection of the tangents to $B$ and $D$ lies on the $A$-symmedian of $\triangle BAD$. Thus, since $A, C$, and this point are collinear, we must have $C$ lying on the $A$-symmedian. Claim: $\frac{BC}{CD} = \frac{BA}{AD}.$ Proof: Let $X = AC\cap BD$. Then, using a well known property of the symmedian, we have $$\frac{BX}{XD} = \left(\frac{AB}{AD}\right)^2.$$However, by the Ratio Lemma on $\triangle BCD$, we have $$\frac{BX}{XD} = \frac{BC\sin \angle BCX}{DC\sin \angle DCX} = \frac{BC\sin \angle BCA}{DC\sin \angle DCA} = \frac{BC\sin \angle BDA}{DC\sin \angle DBA} = \frac{BC}{DC}\cdot \frac{AB}{AD}$$by Inscribed Angles and the Law of Sines on $\triangle ABD$. Thus, we have $$\frac{BC}{DC}\cdot \frac{AB}{AD} = \left(\frac{AB}{AD}\right)^2\iff \frac{BC}{CD} = \frac{AB}{AD},$$as desired. $\square$ Now, by Steiner's (I think that's what it's called?) on $\triangle ABD$, we have $$\frac{BM}{MD}\cdot \frac{BX}{XD} = \left(\frac{AB}{AD}\right)^2,$$as $M$ and $X$ are isogonal with respect to $\triangle ABD$ and on line $BD$. However, since $\frac{AB}{AD} = \frac{CB}{CD}$, by the converse of Steiner's (which follows from Steiner's and the fact that $f(P) = \frac{BP}{PD}$ is injective as $P$ varies on line $BD$), we have that $M$ and $X$ are isogonal with respect to $\triangle BCD$. Now, this means that $$\angle CDM = \angle CDB = \angle CAB = \angle XAB = \angle MAD$$due to Inscribed Angles, the fact that $A, X$, and $C$ are collinear, and the isogonality (is that even a word?). Furthermore, we have $$\angle MCD = \angle XCB = \angle ACB = \angle ADB = \angle ADM$$for the same reasons, so we have that $$\angle CMD = 180 - \angle MCD - \angle CDM = 180 - \angle ADM - \angle MAD = \angle AMD,$$as desired.