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We are tasked with proving that \( OX = OY \), where \( O \) is the point of intersection of the diagonals of the parallelogram \( ABCD \), and a line through \( O \) meets sides \( AB \) and \( CD \) at points \( X \) and \( Y \), respectively.

### Step-by-step proof:

1. **Diagonals of a Parallelogram Bisect Each Other:**
   Since \( ABCD \) is a parallelogram, its diagonals bisect each other at \( O \). This means that \( OA = OC \) and \( OB = OD \). This property will be useful in proving the equality of the segments \( OX \) and \( OY \).

2. **Line through \( O \) Intersects \( AB \) and \( CD \):**
   Let the line passing through \( O \) intersect sides \( AB \) and \( CD \) at points \( X \) and \( Y \), respectively.

3. **Using Similar Triangles:**
   We will now look at triangles \( OAX \) and \( OCY \):
   - Both triangles share the angle at \( O \).
   - \( \angle OAX = \angle OCY \) because they are alternate interior angles formed by the transversal \( OX \) (or \( OY \)) and the parallel sides \( AB \parallel CD \) (since opposite sides of a parallelogram are parallel).
   - By the same reasoning, \( \angle OXA = \angle OYC \).

   Hence, triangles \( OAX \) and \( OCY \) are similar by **AA similarity**.

4. **Proportionality of Corresponding Sides:**
   From the similarity of triangles \( OAX \) and \( OCY \), we have the following proportionality:
   \[
   \frac{OX}{OA} = \frac{OY}{OC}
   \]
   Since \( OA = OC \) (as diagonals of a parallelogram bisect each other), we can conclude:
   \[
   OX = OY
   \]

Thus, we have proven that \( OX = OY \).

Would you like more details or further clarification? Here are some related questions for further exploration:

1. How would the proof change if the shape were not a parallelogram?
2. What is the significance of the diagonals bisecting each other in this proof?
3. Can we use coordinate geometry to prove the same result?
4. What if the line through \( O \) intersects other sides of the parallelogram? Would the result still hold?
5. What happens if the parallelogram is a rhombus or a rectangle? Does the equality of \( OX \) and \( OY \) hold?

**Tip:** Always check for properties like bisecting diagonals when working with parallelograms, as they often simplify proofs.

The diagonals of the parallelogram ABCD meet at O. A line through O meets the sides AB and CD at X and Y respectively. Prove that OX = OY.

Learn how to prove that OX = OY where O is the intersection of diagonals in parallelogram ABCD. This problem explores geometric properties of parallelograms and uses the AA similarity theorem to establish the solution.

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