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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 $ABD$.

 Sep 23, 2024
 #1
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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. 

 

Thanks! :)

 Sep 24, 2024
edited by NotThatSmart  Sep 24, 2024

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