2013 AMC 10A Problem 12

The problem never says where $E$ is on $BC$. So the perimeter should be the same wherever $E$ is. Here is why.

1. Equal base angles. $AB = AC$, so the triangle is isosceles and the angles at $B$ and $C$ are equal.
2. Look at the small triangle $DBE$. $DE$ is parallel to $AC$, so the angle $DEB$ equals the angle $C$ (corresponding angles). And angle $C$ equals angle $B$. So triangle $DBE$ has two equal angles, at $B$ and at $E$. That makes it isosceles: $DE = DB$.
3. Look at the small triangle $FEC$. In the same way, $EF$ is parallel to $AB$, so angle $FEC$ equals angle $B$, which equals angle $C$. So $EF = FC$.
4. Add up the sides of $ADEF$. Replace $DE$ by $DB$ and $EF$ by $FC$:
$$AD + DE + EF + FA = (AD + DB) + (FC + FA).$$
5. $AD + DB$ is all of side $AB = 28$, and $FC + FA$ is all of side $AC = 28$. The perimeter is $28 + 28 = 56$.

(The length $BC = 20$ is never needed.)

The answer is $\boxed{\textbf{(C)}\ 56}$.

Lines parallel to the sides of an isosceles triangle cut off smaller isosceles triangles. That lets you trade a slanted piece for an equal piece lying along a side. When a point can move freely and the question still has one answer, look for a trade like this.

Since $E$ can be anywhere, put it at the middle of $BC$. Then $DE$ and $EF$ run from the middle of $BC$ to the middles of $AB$ and $AC$, so $D$ and $F$ are the midpoints of $AB$ and $AC$. Each side of $ADEF$ is half of $28$, which is $14$, and the perimeter is $4 \times 14 = 56$.