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 A D C 45
12
Angle ABD = 180 - 2(45) = 90
Then AB = BC = AC / sqrt 2 = 12 / sqrt 2 = 6sqrt 2
The area = (1/2) (6 sqrt 2)^2 = 18 * 2 = 36
+1459 
