Answer these questions step-by-step the way you would if you were answering a test paper 

Certainly! I'll guide you step-by-step through answering each question, as you might on a test paper.

### Question 4.1: Prove that \(\frac{PE}{EB} = \frac{PF}{FC}\)

**Step 1: Understand the Problem**

- You are given a parallelogram \(ABCD\) with \(P\) on line \(DB\) extended and \(PC\) extended intersecting line \(AD\) at point \(Q\).
- The task is to prove the ratio of segments \(PE\) to \(EB\) equals the ratio of segments \(PF\) to \(FC\).

**Step 2: Analyze the Parallelogram Properties**

- Recall that opposite sides of a parallelogram are parallel and equal in length: \(AB \parallel CD\) and \(BC \parallel AD\).
- The diagonals of a parallelogram bisect each other: \(DB\) and \(AC\) intersect at their midpoint.

**Step 3: Identify Similar Triangles**

- Consider triangles \(\triangle PEB\) and \(\triangle PFC\).
- Since \(AB \parallel CD\) and \(DB\) intersects these parallel lines, angles \(\angle PEB\) and \(\angle PFC\) are corresponding angles.
- Also, \(\angle PBE = \angle PFC\) as vertical angles.

**Step 4: Prove Triangle Similarity**

- From the angle relationships, \(\triangle PEB \sim \triangle PFC\) by AA (Angle-Angle) similarity criterion.

**Step 5: Use Proportionality of Similar Triangles**

- By similarity, corresponding sides of similar triangles are proportional.
- Thus, \(\frac{PE}{EB} = \frac{PF}{FC}\).

**Conclusion**

- The proof is complete by showing that triangles \(\triangle PEB\) and \(\triangle PFC\) are similar, establishing the required ratio.

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### Question 4.2: Prove that \(\frac{PN}{NB} = \frac{PQ}{QC}\)

**Step 1: Understand the Problem**

- You need to prove the ratio of segments \(PN\) to \(NB\) equals the ratio of segments \(PQ\) to \(QC\).

**Step 2: Use Parallel Line and Transversal Properties**

- As \(ABCD\) is a parallelogram, \(AD \parallel BC\), and \(DB\) serves as a transversal intersecting them.
- Similarly, consider the line \(PC\) extended, intersecting \(AD\) at \(Q\).

**Step 3: Identify and Prove Similar Triangles**

- Consider triangles \(\triangle PNB\) and \(\triangle PQC\).
- Since \(AD \parallel BC\), angles \(\angle PNB\) and \(\angle PQC\) are corresponding angles.
- Angles \(\angle PBN\) and \(\angle PCQ\) are vertical angles and hence equal.

**Step 4: Prove Triangle Similarity**

- By AA (Angle-Angle) similarity, \(\triangle PNB \sim \triangle PQC\).

**Step 5: Use Proportionality of Similar Triangles**

- From the similarity of triangles, \(\frac{PN}{NB} = \frac{PQ}{QC}\).

**Conclusion**

- The similarity of \(\triangle PNB\) and \(\triangle PQC\) proves the required proportionality.

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### Additional Information

By following these structured steps, you systematically prove the geometric relationships, leveraging the properties of parallelograms and similar triangles.

If you have any questions or need further clarification, feel free to ask!

Learn how to prove ratios of segments in a parallelogram using similar triangles with step-by-step solutions. This problem involves applying geometric concepts like parallelogram properties and AA similarity criterion. Suitable for high school students.

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