+900 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.
Thus, we can write
AB=BC=12/√2=6√2
Also, let's note that from the problem, we have thart
AD=ADBD=BDAB=BC
This means that triangles ABD and CBD are congruent.
Thus, we have
[ABD]=(1/2)[ABC]=(1/2)(1/2)(6√2)2=(1/4)(72)=18
Thus, our answer is 18.
Thanks! :)
+1950 First, let's note that triangle ABC is a 45-45-90 triangle.
Thus, we can write
AB=BC=12/√2=6√2
Also, let's note that from the problem, we have thart
AD=ADBD=BDAB=BC
This means that triangles ABD and CBD are congruent.
Thus, we have
[ABD]=(1/2)[ABC]=(1/2)(1/2)(6√2)2=(1/4)(72)=18
Thus, our answer is 18.
Thanks! :)
