In triangle $A B C, A B=2, A C=1+\sqrt{5}$, and $\angle C A B=54^{\circ}$. Suppose $D$ lies on the extension of $A C$ through $C$ such that $C D=\sqrt{5}-1$. If $M$ is the midpoint of $B D$, determine the measure of $\angle A C M$, in degrees.
Let $E$ be the midpoint of $\overline{A D} . E C=\sqrt{5}+1-\sqrt{5}=1$, and $E M=1$ by similar triangles $(A B D \sim E M D) . \triangle E C M$ is isosceles, with $m \angle C E M=54^{\circ}$. Thus $m \angle A C M=m \angle E C M=$ $\frac{180-54}{2}=63^{\circ}$. $\fbox{63}$.