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 $ABD$.

onyuIee Aug 14, 2024

#1**+1 **

Let's make some observations of the problem.

First, note that triangle ABC is a 45-45-90 right triangle. This means that \(AB = BC = 12/\sqrt 2 = 6\sqrt 2\)

Also, we have that

\(AD = AD\\ BD = BD \\ AB = BC\)

This means that triangles ABD and CBD are congruent

Thus, we have \([ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18\)

So our aswer is 18.

NotThatSmart Aug 14, 2024