+207 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 $ABD$.
+1908 First, let's note that triangle ABC is a 45-45-90 triangle.
We also now know that \(AB = BC = 12/\sqrt{ 2} = 6\sqrt{ 2} \)
We have that
\(AD = AD \\ BD = BD \\ AB = BC\)
So by SSS congruence, triangles ABd and CBD are congruent.
So, this means we have
\( [ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt{ 2})^2 = (1/4) (72) = 18\)
So 18 is our answer!
Thanks! :)
