In the figure below, not drawn to scale, EFGH is a square overlapping a trapezium ABCD. AB // DC, ∠CHG = 25° and ∠IFC = 33°.

1. Identify the properties of the square EFGH and its diagonal HF

In square $EFGH$, $HF$ is a diagonal. The diagonal of a square bisects the interior angle.

2. Calculate angle x

$\angle EHF = \frac{90^\circ}{2} = 45^\circ$
So, $x = 45^\circ$.

1. Find $\angle HIG$ and $\angle FIC$ using the properties of the square.

Since $EFGH$ is a square, $\angle HGF = 90^\circ$.
Assuming $I$ is the intersection of $FG$ and $DC$, $F, I, G$ are collinear.
In $\triangle HGI$, the sum of angles is $180^\circ$:
$\angle HIG = 180^\circ - 90^\circ - 25^\circ = 65^\circ$
$\angle FIC = \angle HIG = 65^\circ$ (vertically opposite angles)

2. Find $\angle C$ (or $\angle BCD$) using the angle sum of $\triangle FIC$.

In $\triangle FIC$:
$\angle C = 180^\circ - \angle FIC - \angle IFC$
$\angle C = 180^\circ - 65^\circ - 33^\circ = 82^\circ$

3. Find $\angle EBF$ using interior angles between parallel lines.

Since $AB \parallel DC$, the interior angles on the same side of the transversal $BC$ sum to $180^\circ$.
$\angle EBF + \angle C = 180^\circ$
$\angle EBF = 180^\circ - 82^\circ = 98^\circ$