+0  
 
0
4
1
avatar+357 

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 $ABD$.

 Mar 27, 2024
 #1
avatar+128794 
+1

A

 

             D       12

 

B                 C

 

Triangle ABC is  a 45-45-90 right triangle

AB = BC  =   12/sqrt 2 =  6sqrt 2

 

AD = AD

BD = BD

AB = BC

So triangles  ABD  and CBD  are congruent

 

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

 

cool cool cool

 Mar 27, 2024

0 Online Users