2021 AMC Fall 10A Problem 7

The plan is to find the angles of triangle $DEF$, then use the straight line $AD$ to get $\angle AFE$.

1. Angles around $D$. All the angles around a point add up to $360^\circ$. At $D$ there are three of them:
- $\angle ADC = 90^\circ$, because $ABCD$ is a square.
- $\angle CDE = 110^\circ$, given.
- $\angle EDA$, the one we want.

Because $E$ is on the other side of line $CD$ from $A$ (see the figure), these three do not overlap. So
$$\angle EDA = 360^\circ - 90^\circ - 110^\circ = 160^\circ.$$
2. Angle at $D$ in the triangle. $F$ is on segment $AD$, so $\angle FDE$ is the same angle: $160^\circ$.
3. The other two angles of triangle $DEF$. Since $DE = DF$, the triangle has two equal sides, so the angles across from them are equal: $\angle DFE = \angle DEF$. The three angles add to $180^\circ$, so each of these is
$$\frac{180^\circ - 160^\circ}{2} = 10^\circ.$$
4. Use the straight line. $A$, $F$, $D$ lie on one line with $F$ in the middle. The angles $\angle AFE$ and $\angle DFE$ sit side by side on that line, so they add to $180^\circ$:
$$\angle AFE = 180^\circ - 10^\circ = 170^\circ.$$

The answer is $\boxed{\textbf{(D)}\ 170}$.

This is a chain of three basic angle facts: angles around a point add to $360^\circ$, angles in a triangle add to $180^\circ$ (with equal angles across from equal sides), and angles along a straight line add to $180^\circ$. The only place to be careful is the picture: the phrase "opposite half-plane" tells you the $110^\circ$ angle opens away from the square, so it adds to the $90^\circ$ instead of overlapping it.