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 $ABD$.
A
D 12
B C
Triangle ABC is a 45-45-90 right triangle
AB = BC = 12/sqrt 2 = 6sqrt 2
Angle ABD = angle CBD
BD = BD
So, by SAS, triangles ABD and CBD are congruent
Then [ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6sqrt 2)^2 = (1/4) (72) = 18
+359 
