7. [25] Point $P$ lies inside a convex pentagon $A F Q D C$ such that $F P D Q$ is a parallelogram. Given that $\angle F A Q=\angle P A C=10^{\circ}$, and $\angle P F A=\angle P D C=15^{\circ}$. What is $\angle A Q C$ ?
Answer: $\frac{\pi}{12}$ Let $C^{\prime}$ be the point such that there is a spiral similarity between $\triangle A F P$ and $\triangle A Q C^{\prime}$. In other words, one triangle can be formed from the other by dilating and rotating about one of the triangle's vertices (in this case, $A$ ). We will show that $C^{\prime}$ is $C$, so our answer will be $\angle A Q C=$ $\angle A Q C^{\prime}=\angle A F P=15^{\circ}$. By the spiral similarity theorem, $\triangle A F Q \sim \triangle A P C^{\prime}$ (this is intuitive by looking at a diagram), so $\angle P A C=\angle F A Q=10^{\circ}$, so to show that $C^{\prime}$ is $C$, it is sufficient to show that $\angle P D C^{\prime}=15^{\circ}$.
Let $X$ be the fourth point of the parallelogram $F P C^{\prime} X$ (see the above diagram). The angle between lines $\overline{F P}$ and $\overline{F A}$ is $15^{\circ}$. Since $\overline{X C^{\prime}} \| \overline{F P}$, the angle between $\overline{F A}$ and $\overline{X C^{\prime}}$ is $15^{\circ}$ as well. In addition, the angle between $\overline{Q A}$ and $\overline{Q C^{\prime}}$ is $\angle A Q C^{\prime}=15^{\circ}$, so $\angle X C^{\prime} Q=\angle F A Q$. Further, because $F P C^{\prime} X$ is a parallelogram, $\frac{Q C^{\prime}}{X C^{\prime}}=\frac{Q C^{\prime}}{F P}$. By similar triangles $\triangle A F P$ and $\triangle A Q C^{\prime}, \frac{Q C^{\prime}}{F P}=\frac{Q A}{F A}$. By SAS similarity, there is a spiral similarity between $\triangle X C^{\prime} Q$ and $\triangle F A Q$, so $\angle F Q X=\angle A Q C^{\prime}=15^{\circ}$.
Note that the segments $\overline{F P}, \overline{X C^{\prime}}$, and $\overline{Q D}$ are all parallel and equal in length. Therefore, $\triangle F Q X \cong$ $\triangle P D C^{\prime}$ are congruent, and $\angle P D C^{\prime}=15^{\circ}$ as well. So $C^{\prime}$ is $C$, and $\angle A Q C=15^{\circ}$.