2021 AMC Fall 10A Problem 7

Look at the three angles around vertex $D$. Since $ABCD$ is a square, $\angle CDA = 90^\circ$. We are given $\angle CDE = 110^\circ$, and because $E$ is on the far side of line $CD$ from $A$, the rays $DA$, $DC$, $DE$ go around $D$ in order, so
$$\angle ADE = 360^\circ - 90^\circ - 110^\circ = 160^\circ.$$

Point $F$ lies on segment $DA$, so $\angle FDE = \angle ADE = 160^\circ$. In triangle $DEF$, $DE = DF$ makes the triangle isosceles with apex $D$, so the two base angles are equal:
$$\angle DFE = \angle DEF = \frac{180^\circ - 160^\circ}{2} = 10^\circ.$$

Finally $F$ is between $A$ and $D$, so $\angle AFE$ and $\angle DFE$ are supplementary:
$$\angle AFE = 180^\circ - 10^\circ = 170^\circ.$$

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

Angle chasing is a chain: go around a vertex (sum $360^\circ$), inside a triangle (sum $180^\circ$, equal base angles when two sides are equal), and along a line (sum $180^\circ$). The only judgment call is orientation; the phrase "opposite half-plane" tells you the $110^\circ$ angle opens away from the square, so it and the right angle add rather than overlap.

• Computing angle FDE as 110 - 90 = 20 or as 180 - 110 = 70 instead of 360 - 90 - 110 = 160, since E sits across line CD from A.
• Reporting $\angle DFE = 10^\circ$ or its complement instead of the supplementary angle $\angle AFE$.