+1678 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.
+129771 A
4 5
B D C
3
Let DC = x and BD = 3 - x
Since AD is an angle bisector
DC / AC = BD / AB
x / 5 = (3 -x) / 4 cross-multiply
5 (3 - x) = 4x
15 - 5x = 4x
15 = 9x
x = 15/9 = 5/3 = DC = the base of [A
And the height of triagle ADC = AB = 4
So
[ADC ] = (1/2) DC * AB = (1/2)(5/3)(4) = 20 / 6 = 10 / 3 = 3 (to the nearest integer)
