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 $ABD$.

MeIdHunter Jul 24, 2024

#1**+1 **

Let's make some observations about the problem.

First off, let's note that Triangle ABC is a 45-45-90 right triangle

This means that we have \( AB = BC = 12/\sqrt 2 = 6\sqrt 2\)

From the problem, we also have that

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

So triangles ABD and CBD are congruent

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

Thanks! :)

NotThatSmart Jul 24, 2024