+972 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 define some variables to help solve this problem.
Let's let BD=3−x and CD=x
From the problem and requirements given, we can write the equation
BD/AB=CD/AC
Plugging in our values from the problem and x, we have the equation
(3−x)/4=x/55(3−x)=4x15−5x=4x15=9xx=15/9=5/3=CD
Now, we can find the area of ADC easily! We have
[ADC]=(1/2)(CD)(AB)=(1/2)(5/3)(4)=10/3
So 10/3 is our answer.
Thanks! :)
