+1439 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 $ABD$.
+1926 Let's make some careful observations about the problem.
First, note that Triangle ABC is a 45-45-90 right triangle
From this, we can conclude that
\(AB = BC = 12/\sqrt 2 = 6\sqrt 2 \)
From the problem, we also know that
\(AD = AD \\ BD = BD \\ AB = BC\)
So triangles ABD and CBD are congruent
Thus, we have that \([ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18\)
So 18 is the answer.
Thanks! :)
