Given the diagram and the information that:

• $AFGB$ is a straight line,
• $BCEF$ is a parallelogram,
• Triangle $EFG$ is isosceles,
• Angles $\angle AFE = 40^\circ$ and $\angle AEB = 65^\circ$,

We need to find:

1. $\angle GFE$
2. $\angle FEA$

(i) Finding $\angle GFE$:

Since triangle $EFG$ is isosceles, the base angles $\angle GFE$ and $\angle FEG$ are equal. Thus, let $\angle GFE = \angle FEG = x$.

In a triangle, the sum of the interior angles is always $180^\circ$. Therefore, we can write the equation for triangle $EFG$:

$x + x + \angle GEF = 180^\circ$

We already know $\angle GEF = 65^\circ$ (given in the diagram). Substituting this into the equation:

$2x + 65^\circ = 180^\circ$

$2x = 180^\circ - 65^\circ$

$2x = 115^\circ$

$x = \frac{115^\circ}{2} = 57.5^\circ$

Thus, $\angle GFE = 57.5^\circ$.

(ii) Finding $\angle FEA$:

We are given that $AFGB$ is a straight line, so:

$\angle AFE + \angle GFE = 180^\circ$

Substituting the values we know:

$40^\circ + \angle FEA = 180^\circ$

$\angle FEA = 180^\circ - 40^\circ - 57.5^\circ = 82.5^\circ$

Thus, $\angle FEA = 82.5^\circ$.

Summary of Results:

1. $\angle GFE = 57.5^\circ$
2. $\angle FEA = 82.5^\circ$

Would you like more details or have further questions?

Here are 5 relative questions for deeper understanding:

1. How do we determine the angles in an isosceles triangle?
2. Why is the sum of the interior angles of a triangle always $180^\circ$?
3. How does the property of straight lines help in angle calculation?
4. What other properties of parallelograms can help solve similar problems?
5. How do angle relationships change in different types of polygons?

Tip: Always use known properties (e.g., straight lines and triangle angle sums) to break down more complex geometric shapes.