The figure below is not drawn to scale. ABE is an equilateral triangle and BCD is an isosceles triangle. ABC and BDF are straight lines and AC // EF. ∠EBD = 90° and ∠BED = 43°.

1. Find $\angle ABE$ using the properties of an equilateral triangle.

Since triangle $ABE$ is equilateral,
$\angle ABE = 60^\circ$

2. Find $\angle CBF$ using vertically opposite angles.

Since $ABC$ and $FBD$ are straight lines, $\angle CBF$ and $\angle ABD$ are vertically opposite angles.
$\angle ABD = \angle ABE + \angle EBD$
$\angle ABD = 60^\circ + 90^\circ = 150^\circ$
Therefore, $\angle CBF = \angle ABD = 150^\circ$

1. Find $\angle EDB$ using the sum of angles in $\triangle EBD$

In $\triangle EBD$, the sum of angles is $180^\circ$.
$\angle EDB = 180^\circ - \angle EBD - \angle BED$
$\angle EDB = 180^\circ - 90^\circ - 43^\circ = 47^\circ$

2. Find $\angle EDF$ using angles on a straight line

Since $BDF$ is a straight line, $\angle EDF$ and $\angle EDB$ are supplementary.
$\angle EDF = 180^\circ - \angle EDB$
$\angle EDF = 180^\circ - 47^\circ = 133^\circ$