+478 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 Let's make some observations about the problem.
First off, let's note that Triangle ABC is a 45-45-90 right triangle
This means that we have AB=BC=12/√2=6√2
From the problem, we also have that
AD=ADBD=BDAB=BC
So triangles ABD and CBD are congruent
Thus, we can conclude that we have the equation[ABD]=(1/2)[ABC]=(1/2)(1/2)(6√2)2=(1/4)(72)=18
Thanks! :)
