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 $ABD$.
B
45 45
A 45 D 45 C
Triangles ABD and CBD are congruent....BD is the altitude of both
AD = BD = CD = 6
[ ABD ] = (1/2) AD * BD = (1/2) (6)(6) = 18
+753 
