+216 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 $ABD$.
+1950 First, let's note that triangle ABC is a 45-45-90 triangle.
We also now know that AB=BC=12/√2=6√2
We have that
AD=ADBD=BDAB=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√2)2=(1/4)(72)=18
So 18 is our answer!
Thanks! :)
