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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 $ABD$.

 Jan 27, 2025
 #1
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Angle B  =90

 

So ABC is right

 

The hypotenuse,  AC  = 12

 

So  AB  , CB  =   12/sqrt 2

 

AD = BD

 

[ ABD ]  = [ ABC ]  / 2  = (1/2) (1/2) (12/sqrt 2)^2  = (1/4) (144) / 2 =  18

 

 

cool cool cool

 Jan 27, 2025

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