2016 AMC 12A Problem 12

Problem 12 of 25IntermediateAlgebra Geometry

In

\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?

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Solution

Applying the Angle Bisector Theorem to

\triangle ABC

gives

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

so

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

Now

BF

lies along the bisector of

\angle ABD

in

\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.

More practice

Concepts: angle bisector theorem · ratio and proportion

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.