ABCD and CFGE are identical parallelograms. HCB and EDC are identical right-angled triangles. (a) Find ∠w. (b) Find ∠y. (c) Circle the words that describe BHF in the statement: BHF (is / is not) a straight line because the sum of angles BHC and CHF (is / is not) 180°.

In parallelogram ABCD, opposite angles are equal, and consecutive angles sum to $180^\circ$.
From the diagram, $\angle ADC = 80^\circ$.
Therefore, $\angle BCD = 180^\circ - \angle ADC = 180^\circ - 80^\circ = 100^\circ$.

The problem states that ABCD and CFGE are identical parallelograms. This means their corresponding angles are equal.
Angle $w$ is $\angle FGE$. This angle corresponds to $\angle BCD$ in parallelogram ABCD.
So, $\angle FGE = \angle BCD = 100^\circ$.
Alternatively, the diagram shows $\angle CGF = 80^\circ$. In parallelogram CFGE, $\angle FGE$ and $\angle CGF$ are consecutive angles, so they sum to $180^\circ$.

The value of $\angle w$ is $\boxed{100^\circ}$.

From Part (a), $\angle BCD = 100^\circ$.
Since CFGE is identical to ABCD, its corresponding angles are equal. The angle $\angle FCE$ corresponds to $\angle DAB$ in parallelogram ABCD.
In parallelogram ABCD, $\angle DAB = 180^\circ - \angle ABC = 180^\circ - 80^\circ = 100^\circ$.
So, $\angle FCE = 100^\circ$.

The problem states that HCB and EDC are identical right-angled triangles.
In $\triangle HCB$: $\angle CHB = 90^\circ$ (indicated by the square symbol) and $\angle HBC = 23^\circ$.
The sum of angles in a triangle is $180^\circ$.

In $\triangle EDC$: $\angle CDE = 90^\circ$ (indicated by the square symbol) and $\angle DEC = 23^\circ$.

From the diagram, the angles $\angle HCB$, $\angle HCF$, and $\angle FCE$ are arranged around point C.
We can find $\angle HCD$ by subtracting $\angle HCB$ from $\angle BCD$:

Similarly, we can find $\angle DCF$ by subtracting $\angle DCE$ from $\angle FCE$:

Angle $y$ is the sum of $\angle HCD$ and $\angle DCF$:

The value of $\angle y$ is $\boxed{66^\circ}$.

For BHF to be a straight line, the sum of angles $\angle BHC$ and $\angle CHF$ must be $180^\circ$.
From the diagram, $\angle BHC = 90^\circ$.

Consider $\triangle HCF$.
We found $\angle HCF = y = 66^\circ$.
From the diagram, $\angle HFC = 25^\circ$.
The sum of angles in $\triangle HCF$ is $180^\circ$.

Since $179^\circ \neq 180^\circ$, BHF is not a straight line.

The completed statement is:
BHF (is / is not) a straight line because the sum of angles BHC and CHF (is / is not) $180^\circ$.