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 $ABD$.
45 B 45
A 45 D 45 C
The triangle is isosceles AB = BC = 12/2 = 6
Angle ABC = 90
So....Angle ABD = 45
So BD = AD = 6
Area ABD = (1/2) (AD) ( BD) = (1/2) (6)(6) = 18
+688 
