+1224 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 $ABD$.
+1952 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/√2=6√2
Also, we have the relation of
AD=ADBD=BDAB=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√2)2=(1/4)(72)=18
So 18 is our final answer.
Thanks! :)
