图中所示为一个圆心为 $O$、直径为 $AB$ 的圆。点 $D$ 和 $F$ 在圆上。点 $E$ 使得 $EB$ 和 $EF$ 是圆的切线。

1. Identify the angle between the diameter and the tangent.

Since $AB$ is a diameter and $EB$ is a tangent to the circle at $B$,
$\angle ABE = 90^\circ$ (tangent $\perp$ radius).

2. Apply the Midpoint Theorem to triangle ABE.

In $\triangle ABE$, $D$ is the midpoint of $AE$ and $C$ is the midpoint of $BE$.
By the Midpoint Theorem, $CD \parallel AB$.

3. Use parallel lines to find angle DCE.

Since $CD \parallel AB$ and $EB$ is a transversal,
$\angle DCE = \angle ABE$ (corresponding angles).
Therefore, $\angle DCE = 90^\circ$.

1. Identify the angles in the equilateral triangle

Since triangle $BEF$ is equilateral, $\angle EFB = \angle BEF = 60^\circ$.

2. Apply the alternate segment theorem to relate the angles

Since $EF$ is a tangent to the circle at $F$ and $FB$ is a chord, by the alternate segment theorem, the angle between the tangent and the chord is equal to the angle in the alternate segment.
Therefore, $\angle BAF = \angle EFB$.
Since $\angle EFB = \angle BEF$, it follows that $\angle BAF = \angle BEF$.