2013 AMC 10A Problem 12

Since $AB = AC$, the base angles satisfy $\angle B = \angle C$.

Because $DE \parallel AC$, the corresponding angles give $\angle DEB = \angle ACB = \angle C = \angle B$. So triangle $DBE$ has two equal angles at $B$ and $E$, making it isosceles with $DB = DE$.

Likewise $EF \parallel AB$ gives $\angle FEC = \angle ABC = \angle B = \angle C$, so triangle $FEC$ is isosceles with $EF = FC$.

Now add up the parallelogram's sides, replacing $DE$ by $DB$ and $EF$ by $FC$:
$$AD + DE + EF + FA = (AD + DB) + (FC + FA) = AB + AC = 28 + 28 = 56.$$

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

Lines parallel to the sides of an isosceles triangle cut off smaller isosceles triangles, which lets you trade a slanted segment for a segment along a side. The perimeter is then independent of the position of $E$; whenever a problem gives a movable point and asks for a fixed quantity, expect a trade like this. The length $BC = 20$ is never used.

Since the answer cannot depend on $E$, put $E$ at the midpoint of $BC$. Then $D$ and $F$ are the midpoints of $AB$ and $AC$ (midline theorem), and all four sides of $ADEF$ equal $14$, so the perimeter is $56$.

• Trying to locate E and compute lengths with similar-triangle ratios, when the answer does not depend on where E is.
• Using $BC = 20$ somewhere, for example answering $28 + 20 = 48$.