In triangle $ABC,$ the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D.$ If $\angle BAC = 60^\circ,$ $\angle ABC = 30^\circ,$ and $AD = 24,$ then find the area of triangle $ABC.$
Consider the altitude of the triangle dropped from $D$ to $\overline{AB},$ as shown below.
[asy]
unitsize(1.5 cm);
pair A, B, C, D, H;
B = (0,0);
C = (2,0);
A = extension(B, B + (dir(30)*(1,0)), C, C + (-dir(60)*(1,0)));
D = extension(A, A + dir(210 - 30), B, C);
H = (D + reflect(B,C)*(D))/2;
draw(A--B--C--cycle);
draw(A--D);
draw(D--H,dashed);
label("$A$", A, A - dir(120));
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$H$", H, N);
[/asy]
Let $H$ be the foot of the altitude. Since triangle $ABC$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, $AC = BC/2$ and $HC = AC/2 = BC/4.$
From the Angle Bisector Theorem applied to triangle $ABC,$
\[\frac{BD}{CD} = \frac{AB}{AC} = 2.\]Then $\frac{CD}{BD} = \frac{1}{2}.$ Then
\[\frac{CH}{BH} = \frac{CD}{BD} = \frac{1}{2}.\]Let $BH = x.$ Then $CH = 2x,$ so $HC = 4x.$ By Pythagoras on triangle $HBC,$
\[x^2 + (4x)^2 = (5x)^2,\]which gives us $x = \frac{AB}{10}$ and $CH = \frac{2AB}{5}.$ But $AB = AC = \frac{BC}{2},$ so $HC = \frac{BC}{5}.$
Hence, $HD = DC - HC = \frac{4BC}{5} - \frac{BC}{5} = \frac{3BC}{5}.$ The area of triangle $ABC$ is then
\[\frac{1}{2} \cdot AC \cdot HD = \frac{1}{2} \cdot \left( \frac{BC}{2} \right) \cdot \frac{3BC}{5} = \boxed{\frac{3BC^2}{20}}.\]
Since the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D,$ we know that $AD$ bisects $\angle BAC.$ Therefore, $\angle BAD = \angle DAC = 30^\circ.$
[asy]
unitsize (2 cm);
pair A, B, C, D;
A = dir(60);
B = dir(90 - 30);
C = dir(-90 + 30);
D = interp(B,C,2/3);
draw(A--B--C--cycle);
draw(A--D);
label("$A$", A, N);
label("$B$", B, NE);
label("$C$", C, SW);
label("$D$", D, SE);
[/asy]
Therefore, triangle $ABC$ is an isosceles triangle with $AB = AC.$ Since $\angle ABC = 30^\circ$ and $\angle BAC = 60^\circ,$ we have $\angle ACB = 180^\circ - \angle ABC - \angle BAC = 90^\circ.$ Thus, triangle $ABC$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, so $AB = AC = \frac{AD}{\sqrt{3}/2} = \frac{48}{\sqrt{3}} = 16 \sqrt{3}.$
[asy]
unitsize (2 cm);
pair A, B, C, D;
A = dir(60);
B = dir(90 - 30);
C = dir(-90 + 30);
D = interp(B,C,2/3);
draw(A--B--C--cycle);
draw(A--D);
label("$A$", A, NW);
label("$B$", B, NE);
label("$C$", C, SW);
label("$D$", D, S);
label("$30^\circ$", (0.5,0.1));
label("$30^\circ$", (A + D)/2, E);
label("$60^\circ$", (0.7,0.4));
[/asy]
The area of triangle $ABC$ is given by
\[[ABC] = \frac{1}{2} \cdot AB \cdot AC = \frac{1}{2} \cdot \frac{48}{\sqrt{3}} \cdot \frac{48}{\sqrt{3}} = \boxed{288}.\]