# 2016 AMC 12A Problems/Problem 12

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Solution

By the angle bisector theorem, $\frac{AB}{AE} = \frac{CB}{CE}$

$\frac{6}{AE} = \frac{7}{8 - AE}$, so $AE = \frac{48}{13}$

Similarly, $CD = 4$

Now, we use mass points.

Assign point $C$ a mass of $1$.

Because $\frac{AE}/{EC} = \frac{6}{7}, A$ will have a mass of $\frac{7}{6}$.

Similarly, $B$ will have a mass of $\frac{4}{3}$.

$mE = mA + mC = \frac{13}{6}$.

Similarly, $mD = mC + mB = \frac{7}{3}$.

The mass of $F$ is the sum of the masses of $E$ and $B$.

$mF = mE + mB = \frac{7}{2}$.

This can be checked with $mD + mA$, which is also $\frac{7}{2}$.

So $\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}$