The problem states that the bunting flags are isosceles triangles. Given the angle $EDF = 66^{\circ}$, we need to find the size of angle $DFE$.

1. Problem Description

The problem states that the bunting flags are isosceles triangles. Given the angle $EDF = 66^{\circ}$, we need to find the size of angle $DFE$.

2. Solution Steps

Since the triangle $DEF$ is an isosceles triangle and the sides $DF$ and $EF$ are equal in length (as indicated by the markings on the sides), the angles opposite these sides are equal. Thus, $\angle DEF = \angle EDF = 66^{\circ}$.

The sum of the angles in any triangle is $180^{\circ}$.

Therefore, $\angle EDF + \angle DEF + \angle DFE = 180^{\circ}$.

Substituting the values we know, we get $66^{\circ} + 66^{\circ} + \angle DFE = 180^{\circ}$.

$132^{\circ} + \angle DFE = 180^{\circ}$.

Subtracting $132^{\circ}$ from both sides, we get $\angle DFE = 180^{\circ} - 132^{\circ}$.

$\angle DFE = 48^{\circ}$.

3. Final Answer

The size of angle $DFE$ is $48^{\circ}$.