ACE 和 BDF 是等边三角形。AF = FE 且 BC = CD。求图中的 $\angle y$。

1. Identify the properties of the line segment FC in the equilateral triangle ACE.

In the equilateral triangle $ACE$, $F$ is the midpoint of $AE$ ($AF = FE$). Therefore, the line segment $CF$ is the median to the base $AE$. In an equilateral triangle, the median also bisects the opposite angle. Thus, $\angle FCE = \frac{60^\circ}{2} = 30^\circ$.

2. Identify the properties of the line segment FC in the equilateral triangle BDF.

Similarly, in the equilateral triangle $BDF$, $C$ is the midpoint of $BD$ ($BC = CD$). The line segment $FC$ is the median to the base $BD$, which means it bisects $\angle BFD$. Thus, $\angle DFC = \frac{60^\circ}{2} = 30^\circ$.

3. Calculate the angle $y$ using the sum of angles in a triangle.

Let the intersection of $FD$ and $CE$ be $G$. In $\triangle FGC$, we have $\angle GFC = 30^\circ$ and $\angle GCF = 30^\circ$.
The angle $y$ is the angle facing upwards between the intersecting lines, which corresponds to $\angle FGC$.
$\angle y = 180^\circ - 30^\circ - 30^\circ = 120^\circ$.