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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 $ABD$.

 Aug 12, 2024
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Let's make some careful observations about the problem. 

First, note that Triangle ABC is a 45-45-90 right triangle

From this, we can conclude that

\(AB = BC  =   12/\sqrt 2 =  6\sqrt 2 \)

 

From the problem, we also know that

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

 

So triangles  ABD  and CBD  are congruent

 

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

 

So 18 is the answer. 

 

Thanks! :)

 Aug 13, 2024
edited by NotThatSmart  Aug 13, 2024

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