+788 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, let's note that triangle ABC is a 45-45-90 triangle,
This means that we have the equation \(AB = BC = 12/\sqrt 2 = 6\sqrt 2 \)
Also, we have the relation of
\(AD = AD \\ BD = BD \\ AB = BC\)
Through these equation, we can determine that triangle ABD and triangle CBD are congruent triangles.
Because of this note, we can write the equation
\( [ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18\)
So 18 is our final answer.
Thanks! :)
