+1855 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 Let's set up variables to complete this problem.
Since AD bisects BAC
Lets let BD = 3 - x
Lets let CD = x
Now, we can easily solve the problem with an equation. We have
BD/AB=CD/AC(3−x)/4=x/55(3−x)=4x15−5x=4x15=9xx=15/9=5/3=CD
Thus, taking this value of CD and plugging it into the area equation, we find
[ADC]=(1/2)(CD)(AB)=(1/2)(5/3)(4)=10/3
thus, The answer is 10/3
Thanks! :)
