+339 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 We can define some variables from the information given from the problem.
First, since we know that AD bisects BAC, let's name variables
Let BD = 3 - x
Let CD = x
Thus, we can plug the two numbers in to get the equation
BD/AB=CD/AC(3−x)/4=x/55(3−x)=4x15−5x=4x15=9xx=15/9=5/3=CD
Finally, plugging in these values for the area, we find that
[ADC]=(1/2)(CD)(AB)=(1/2)(5/3)(4)=10/3
Thus, 10/3 is our answer.
Thanks! :)
