In triangle $ABC$, point $D$ is on side $\overline{AC}$ such that line segment $\overline{BD}$ bisects $\angle ABC$. If $\angle A = 45^\circ$, $\angle C = 45^\circ$, and $AC = 12$, then find the area of triangle $ABD$.
Angle B =90
So ABC is right
The hypotenuse, AC = 12
So AB , CB = 12/sqrt 2
AD = BD
[ ABD ] = [ ABC ] / 2 = (1/2) (1/2) (12/sqrt 2)^2 = (1/4) (144) / 2 = 18