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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 2, 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 triangleThis means thatAB=BC=12/2=62

Also, we have that

 

AD=ADBD=BDAB=BC

 

This means that triangles  ABD  and CBD  are congruent

 

Thus, we have [ABD]=(1/2)[ABC]=(1/2)(1/2)(62)2=(1/4)(72)=18

 

So our aswer is 18. 

 

Thanks! :)

 Aug 2, 2024
edited by NotThatSmart  Aug 2, 2024

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