To prove that a quadrilateral is a rhombus, we must show that all its sides are of equal length. Here's the proof using analytical geometry, based on the given conditions that the diagonals bisect each other and are perpendicular.

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Given:

• A quadrilateral $ABCD$ with diagonals $AC$ and $BD$.
• The diagonals bisect each other, i.e., their intersection point $O$ is the midpoint of both $AC$ and $BD$.
• The diagonals are perpendicular to each other.

Proof:

1. Set up coordinates for vertices: Let the coordinates of the vertices of the quadrilateral be:

   • $A(x_1, y_1)$
   • $B(x_2, y_2)$
   • $C(x_3, y_3)$
   • $D(x_4, y_4)$

   Since the diagonals bisect each other, the intersection point $O$ of the diagonals is the midpoint of $AC$ and $BD$: $O\left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right) = \left(\frac{x_2 + x_4}{2}, \frac{y_2 + y_4}{2}\right).$ This ensures that $O$ is common to both diagonals.

2. Condition of perpendicular diagonals: The diagonals $AC$ and $BD$ are perpendicular. The slopes of the diagonals satisfy: $\text{slope of } AC \cdot \text{slope of } BD = -1.$

   • Slope of $AC = \frac{y_3 - y_1}{x_3 - x_1}$,
   • Slope of $BD = \frac{y_4 - y_2}{x_4 - x_2}$.

   Substituting these, we get: $\frac{y_3 - y_1}{x_3 - x_1} \cdot \frac{y_4 - y_2}{x_4 - x_2} = -1.$

3. Distance formula to prove equal sides: To show $ABCD$ is a rhombus, we need to prove: $AB = BC = CD = DA.$ Using the distance formula: $AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, \quad BC = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2},$ $CD = \sqrt{(x_4 - x_3)^2 + (y_4 - y_3)^2}, \quad DA = \sqrt{(x_1 - x_4)^2 + (y_1 - y_4)^2}.$

   From the given conditions:

   • The diagonals bisect each other, implying symmetry about $O$.
   • The diagonals are perpendicular, forming two congruent right triangles for each diagonal (e.g., $\triangle AOB \cong \triangle COB$).

   Therefore, the symmetry and perpendicularity imply that each side of the quadrilateral must be of equal length: $AB = BC = CD = DA.$

4. Conclusion: Since all four sides are equal, $ABCD$ is a rhombus.

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Summary:

We proved that if the diagonals of a quadrilateral bisect each other and are perpendicular, then the quadrilateral is a rhombus, as the perpendicularity and midpoint conditions enforce equal side lengths.

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Would you like further clarification or diagrams to illustrate this proof? 😊

Questions to Explore:

1. How can we generalize this proof to three dimensions?
2. What properties of parallelograms are used in this proof?
3. Can the condition of perpendicular diagonals alone prove that $ABCD$ is a rhombus?
4. How can we use vector methods to prove the same statement?
5. What additional conditions might guarantee a quadrilateral is a square instead of just a rhombus?

Tip:

When working with analytical geometry proofs, always leverage midpoint and slope conditions to confirm bisecting and perpendicularity properties efficiently.