+685 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 $ABD$.
+953 First, let's note that triangle ABC is a 45-45-90 triangle.
Thus, we can write
\(AB = BC = 12/\sqrt 2 = 6\sqrt 2 \)
Also, let's note that from the problem, we have thart
\(AD = AD\\ BD = BD\\ AB = BC\)
This means that triangles ABD and CBD are congruent.
Thus, we have
\([ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18\)
Thus, our answer is 18.
Thanks! :)
+953 First, let's note that triangle ABC is a 45-45-90 triangle.
Thus, we can write
\(AB = BC = 12/\sqrt 2 = 6\sqrt 2 \)
Also, let's note that from the problem, we have thart
\(AD = AD\\ BD = BD\\ AB = BC\)
This means that triangles ABD and CBD are congruent.
Thus, we have
\([ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18\)
Thus, our answer is 18.
Thanks! :)
