In triangle $ABC$, point $D$ is on side $\overline{BC}$ such that line segment $\overline{BD}$ bisects $\angle ABC$. If $\angle A = 45^\circ$, $\angle B = 60^\circ$, and $AC = 12$, then find the area of triangle $ABD$.
By the Law of Sines on triangle $ABC$, \[\frac{AB}{\sin 60^\circ} = \frac{12}{\sin 45^\circ},\]so $AB = 4 \sqrt{6}$.
[asy]
unitsize(0.5 cm);
pair A, B, C, D;
A = (0,0);
C = (12,0);
B = intersectionpoint(arc(A,4*sqrt(6),0,180),arc(C,6,0,180));
D = interp(B,C,1/3);
draw(A--B--C--cycle);
draw(A--D);
label("$A$", A, SW);
dot("$B$", B, NW);
label("$C$", C, SE);
dot("$D$", D, S);
label("$12$", (A + C)/2, S);
label("$4 \sqrt{6}$", (A + B)/2, NW);
[/asy]
Then $\sin \angle ABD = \sin (60^\circ - 45^\circ) = \sin 15^\circ$, so
\[[ABD] = \frac{1}{2} \cdot AD \cdot BD \cdot \sin 15^\circ.\]Then
\[\sin 15^\circ = \sin (45^\circ - 30^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4},\]so
\[[ABD] = \frac{AD \cdot BD (\sqrt{6} - \sqrt{2})}{8}.\]By the Angle Bisector Theorem, $\frac{BD}{DC} = \frac{AB}{AC} = \frac{4 \sqrt{6}}{12} = \frac{\sqrt{6}}{3}$, so $\frac{BD}{12 - BD} = \frac{\sqrt{6}}{3}$.
Cross-multiplying, we get $BD^2 = 6 - 12 BD$, so $BD^2 + 12 BD - 6 = 0$. This quadratic factors as $(BD - 3)(BD + 2) = 0$, so $BD = 3$. Therefore, $AC = 12$, so $CD = AC - AD = 9$, and
\[[ABD] = \boxed{\frac{9 (\sqrt{6} - \sqrt{2})}{8}}.\]