+277 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 $ABD$.
+1892 Let's make some observations about the problem.
First off, let's note that Triangle ABC is a 45-45-90 right triangle
This means that we have \( AB = BC = 12/\sqrt 2 = 6\sqrt 2\)
From the problem, we also have that
\(AD = AD \\ BD = BD \\ AB = BC\)
So triangles ABD and CBD are congruent
Thus, we can conclude that we have the equation\([ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18\)
Thanks! :)
