+216 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.
+1950 Since we know AD bisects BAC, we have
BD = 3 - x
CD = x
Now, we can write
BD/AB=CD/AC(3−x)/4=x/55(3−x)=4x15−5x=4x15=9xx=15/9=5/3=CD
This just gives us
[ADC]=(1/2)(CD)(AB)=(1/2)(5/3)(4)=10/3
Thanks! :)
