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.

ABJeIIy Jul 21, 2024

#1**+1 **

Let's set up variables to complete this problem.

Since AD bisects BAC

Lets let BD = 3 - x

Lets let CD = x

Now, we can easily solve the problem with an equation. We have

\(BD / AB = CD / AC\\ (3 - x) / 4 = x / 5\\ 5 (3 - x) = 4x\\ 15 - 5x = 4x \\ 15 = 9x \\ x = 15/9 = 5/3 = CD\)

Thus, taking this value of CD and plugging it into the area equation, we find

\([ ADC ] = (1/2) (CD) ( AB) = (1/2) ( 5/3) ( 4) = 10 / 3 \)

thus, The answer is \(10/3\)

Thanks! :)

NotThatSmart Jul 21, 2024