2016 AMC 12A Problem 12

Below is the professionally curated solution for Problem 12 of the 2016 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 12A solutions, or check the answer key.

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Concepts:angle bisector theorem ratio and proportion

Difficulty rating: 1500

12.

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 $AF:FD?$

$3:2$

$5:3$

$2:1$

$7:3$

$5:2$

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, $AF:FD=AB:BD=6:3=2:1.$

Thus, the correct answer is C.

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