+812 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 $ABD$.
+1786 First, let's make some observations about the problem.
Let's note that triangle ABC is 45-45-90 triangle.
This means that we can easily find some other lengfths. We have
\(AB = BC = 12/\sqrt 2 = 6\sqrt 2\)
First, let's also note that we als have
\( AD = AD \\ BD = BD \\ AB = BC\)
this finally means that we know that triangle ABD and triangle triangle CBD are congruent triangles.
Thus, we can finally we have
\([ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18\)
So 18 is our final answer.
Thanks! :)
