In △ ABC, AB=6, BC=7, and CA=8. Point D lies on BC, and AD bisects ∠ BAC. Point E lies on AC, and BE bisects ∠ ABC. The bisectors intersect at F. What is the ratio AF:FD?

2016 AMC 12A Problem 12

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12.

\triangle ABC,

AB=6,

BC=7,

and

CA=8.

Point

D

lies on

\overline{BC},

and

\overline{AD}

bisects

\angle BAC.

Point

E

lies on

\overline{AC},

and

\overline{BE}

bisects

\angle ABC.

The bisectors intersect at

F.

What is the ratio

AF:FD?

3:2

5:3

2:1

7:3

5:2

Apply the Angle Bisector Theorem in

\triangle ABC

to find

BD

\triangle ABD,

segment

BF

bisects

\angle ABD,

AF:FD=AB:BD

Applying the Angle Bisector Theorem to

\triangle ABC

gives

BD:DC=AB:AC=6:8,

BD=\dfrac{6}{6+8}\cdot 7=3.

Now

BF

lies along the bisector of

\angle ABD

\triangle ABD,

so by the Angle Bisector Theorem again,

\begin{gathered} AF:FD=AB:BD\\ =6:3=2:1. \end{gathered}

Thus, the correct answer is C.