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 $ABD$.
A45
12/sqrt 2 D 12
B 90 12/sqrt 2 C 45
By AAS, triangle ABD congruent to triangle CBD
[ ABC] = (1/2)(12/sqrt 2)^2 = (1/2)(144/2) = 144/4 = 36
[ABD] = (1/2)[ABC] = 18
The area of triangle ABD is 40*sqrt(2) + 12.
+1055 
