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avatar+1352 

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 $ABD$.

 Dec 14, 2023
 #1
avatar+129840 
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                   B

 

 

   45  A        D           C 45

 

                 12

 

Angle ABD =   180 - 2(45)  =  90

 

Then  AB = BC =    AC / sqrt 2  =   12 / sqrt 2 =   6sqrt 2

 

The area  =  (1/2) (6 sqrt 2)^2  =   18 * 2 =   36

 

 

cool cool cool

 Dec 15, 2023

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