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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 16, 2024
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                 45 B 45

 

 

A 45              D                    45 C

 

The triangle is  isosceles  AB = BC =  12/2 = 6

 

Angle ABC  = 90

So....Angle ABD = 45

 

So BD = AD =  6

 

Area ABD =  (1/2) (AD) ( BD)  = (1/2) (6)(6)  =  18

 

cool cool cool

 Jan 16, 2024

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