We are tasked with proving that $\angle BGF = \angle CHF$ in the given quadrilateral setup. Let's work through the proof step-by-step.

Problem Restatement

• We have a quadrilateral $ABCD$ where $AB = CD$.
• Points $E$ and $F$ are the midpoints of $AD$ and $BC$ respectively.
• Lines $BA$ and $CD$ intersect line $FE$ (extended if necessary) at points $G$ and $H$ respectively.
• We are asked to prove that $\angle BGF = \angle CHF$.

Key Observations

1. $E$ and $F$ are midpoints of the sides $AD$ and $BC$ respectively, which means $EF$ is the midsegment of the quadrilateral $ABCD$. A midsegment in a quadrilateral connects the midpoints of two sides, and it is parallel to the other pair of opposite sides.

2. Since $E$ and $F$ are the midpoints of $AD$ and $BC$, and $AB = CD$, the segment $EF$ is parallel to both $AB$ and $CD$. This fact will play an important role in the proof.

Proof

1. Parallelism: Since $EF$ is the midsegment of quadrilateral $ABCD$ and $AB = CD$, we have $EF \parallel AB \quad \text{and} \quad EF \parallel CD.$ This implies that $EF$ is parallel to both $AB$ and $CD$, as established earlier.

2. Intersecting Lines:

   • $G$ is the point where $BA$ intersects $FE$ extended, and
   • $H$ is the point where $CD$ intersects $FE$ extended.

3. Transversals: Since $EF \parallel AB$ and $EF \parallel CD$, lines $BA$ and $CD$ act as transversals to the parallel lines $EF$.

4. Angles formed by Parallel Lines:

   • Consider the transversal $BA$ and the parallel lines $EF$ and $AB$. By the properties of parallel lines, the angle formed between the transversal $BA$ and the line $EF$ at $G$, i.e., $\angle BGF$, is equal to the angle formed between the transversal $CD$ and the line $EF$ at $H$, i.e., $\angle CHF$.
   • This is because these angles are corresponding angles created by the transversal $BA$ (or $CD$) and the parallel lines $EF$ and $AB$ (or $CD$).

Thus, we have: $\angle BGF = \angle CHF.$

Conclusion

By using the properties of parallel lines and corresponding angles, we have shown that $\angle BGF = \angle CHF$, which completes the proof.

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Do you have any questions about the proof or would you like further details on a specific step? Here are five related questions for further practice:

1. What is a midsegment of a quadrilateral, and how does it relate to the opposite sides?
2. How can we prove that $EF$ is parallel to both $AB$ and $CD$?
3. What are the properties of corresponding angles formed by a transversal intersecting two parallel lines?
4. How does the condition $AB = CD$ influence the configuration of the quadrilateral?
5. Can this result be extended to other types of quadrilaterals, such as parallelograms?

Tip: In geometry, corresponding angles are often key in proving properties related to parallel lines, so identifying them early can simplify proofs.