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 $\boxed{144 \sqrt{3}}.$
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 $ABC.$
1 answer