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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 27, 2024

Best Answer 

 #1
avatar+1230 
+1

First, let's note that triangle ABC is a 45-45-90 triangle. 

Thus, we can write

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

 

Also, let's note that from the problem, we have thart

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

 

This means that triangles ABD and CBD are congruent. 

Thus, we have

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

 

Thus, our answer is 18. 

 

Thanks! :)

 Jun 27, 2024
 #1
avatar+1230 
+1
Best Answer

First, let's note that triangle ABC is a 45-45-90 triangle. 

Thus, we can write

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

 

Also, let's note that from the problem, we have thart

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

 

This means that triangles ABD and CBD are congruent. 

Thus, we have

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

 

Thus, our answer is 18. 

 

Thanks! :)

NotThatSmart Jun 27, 2024

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