In triangle $ABC,$ the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D.$ If $\angle BAC = 60^\circ,$ $\angle ABC = 60^\circ,$ and $AD = 24,$ then find the area of triangle $ABC.$
By SAS, triangle $ADB$ is an equilateral triangle, so $BD = AD = 24$ and $AB = 24 \sqrt{3}.$ Furthermore, $\angle BAD = 60^\circ,$ so $\triangle ABD$ is a $30$-$60$-$90$ triangle. Therefore, $AD = 24 = 2 \cdot BD = BC$, so triangle $ABC$ is an equilateral triangle.
[asy]
unitsize(3 cm);
pair A, B, C, D;
A = (0,0);
B = (1,0);
C = rotate(60)*(1,0);
D = interp(B,C,2/3);
draw(A--B--C--cycle);
draw(A--D);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, NE);
[/asy]
Therefore, the area of triangle $ABC$ is $\boxed{144 \sqrt{3}}.$
Let's first draw a diagram to visualize the triangle.
[asy]
pair A,B,C,D;
A=(0,0);
B=(1,0);
D=(1/2,24*sin(pi/3));
C=extension(A,D,B,(3/2,0));
draw(A--B--C--cycle);
draw(A--D);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,S);
label("$D$",D,N);
[/asy]
Since $BD=DC$ and $\angle BAC = \angle ABC,$ we know that triangle $ABC$ is an equilateral triangle. Therefore, $AB=BC=AC=BC.$
Since $\angle ABC=60^\circ,$ we can use the Law of Cosines to find $BC.$
The Law of Cosines states that $c^2=a^2+b^2-2ab\cos C,$ where $a,$ $b,$ and $c$ are side lengths of a triangle and $C$ is the angle opposite side $c.$
In this case, $a=BC,$ $b=24,$ and $C=60^\circ,$ so we have
\[BC^2=24^2+24^2-2(24)(24)\cos 60^\circ.\]
Simplifying this gives
\[BC^2=576+576-576,\]
so $BC^2=576.$
Since $BC>0,$ we can take the positive square root of both sides, which gives $BC=24\sqrt{2}.$
Therefore, the area of triangle $ABC$ is
\[\frac{\sqrt{3}}{4} BC^2=\frac{\sqrt{3}}{4} (24\sqrt{2})^2 = \frac{\sqrt{3}}{4} \cdot 576 \cdot 2 = \boxed{216\sqrt{3}}.\]