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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 $ABD$.

 Jun 9, 2024
 #1
avatar+1334 
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First, let's note that triangle ABC is a 45-45-90 triangle. 

We also now know that \(AB = BC = 12/\sqrt{ 2} = 6\sqrt{ 2} \)

 

We have that

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

 

So by SSS congruence, triangles ABd and CBD are congruent. 

 

So, this means we have

\( [ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt{ 2})^2 = (1/4) (72) = 18\)

 

So 18 is our answer!

 

Thanks! :)

 Jun 10, 2024

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