+1000 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.
+1946 We know that AD bisects BC. Thus, we can set variables to get some common ground here.
Let's let BD=3−x and let CD=x
There's a couple things we can do, but let's the Angle bisector thereom to get
BD/AB=CD/AC
We already set some variables, so subsituting in what we know, we get
(3−x)/4=x/55(3−x)=4x15−5x=4x15=9xx=15/9=5/3=CD
Alright, so now we know what CD is.
Since the area of ABC can be found with CD and AB, we simply have
[ADC]=(1/2)(CD)(AB)=(1/2)(5/3)(4)=10/3
Thus, 10/3 is our final answer.
Thanks! :)
*1900
