+830 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.
+129847 A
4 5
B D C
3
Since AD is a bisector.....Let BD = x and DC =3 - x and we have that
BD/ AB = DC / AC
x / 4 = (3 - x) / 5
5x = 4 (3 - x)
5x = 12 - 4x
9x= 12
x = 4/3 = BD
[ ABD ] = (1/2) (4/3)(4) = 16/6 = 8/3
[ABC ] = (1/2)(4)(3) = 6
[ADC ] = [ ABC ] - [ ABD ] = 6 - 8/3 = 10/3 = 3 + 1/3 = 3 (to the nearest integer )
