+92 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.
+1808 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
\(BD / AB = CD / AC \\ (3 - x) / 4 = x / 5 \\ 5 (3 - x) = 4x\\ 15 - 5x = 4x \\ 15 = 9x \\ x = 15/9 = 5/3 = CD \)
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! :)
