+1000 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 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/√2=6√2
First, let's also note that we als have
AD=ADBD=BDAB=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√2)2=(1/4)(72)=18
So 18 is our final answer.
Thanks! :)
