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
A 45 D 45 C
12
Angle B = 90
BD = BD and is an altitude
And angle ABD = angle CBD
By AAS, triangle ABD is congruent to triangle CBD
AD = 6 = CD
We have that
AD / DB = BD / CD
6 / BD = BD / 6
BD^2 = 36
BD = 6
Area of ABC = (1/2) AC * BD = (1/2) 12 * 6 = 36
+766 
