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In triangle $ABC$, $\angle ABC = 90^\circ$, and $D$ is on side $\overline{BC}$ such that $\overline{AD}$ bisects $\angle BAC$.  If $AB = 4,$ $BC = 3$, and $AC = 5,$ then find the area of $\triangle ADC$. Round your answer to the nearest integer.

 Oct 13, 2023
 #1
avatar+128794 
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Since AD bisects  BAC

Let  BD  = 3 - x

Let  CD =  x

 

So  we have that

 

BD / AB  = CD / AC

(3 - x) / 4  = x / 5

5 (3 - x)  = 4x

15 - 5x = 4x

15 = 9x

x = 15/9  = 5/3 =  CD

 

[ ADC ] = (1/2) (CD) ( AB)  =  (1/2) ( 5/3) ( 4)  =   10 / 3

 

cool cool cool

 Oct 13, 2023

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