1. the following table describes some of the characteristics of some special quadrilaterals.
table with rows (types of quadrilaterals: square, rhombus, rectangle, isosceles trapezoid, kite, etc.) and columns (diagonals bisect each other, diagonals are congruent, diagonals are perpendicular bisectors, etc.) with checkmarks
which of the given conditional statements has no counterexample?
- if a quadrilateral has diagonals that are perpendicular bisectors of each other, then it is a rhombus
- if a quadrilateral has all sides congruent, then it is a square
- if a quadrilateral has at least two angles that are congruent, then it is a trapezoid
- if a quadrilateral has two pairs of congruent sides, then it is a parallelogram

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# Brief Explanations:
To determine which conditional statement has no counterexample, we analyze each:

1. **"If a quadrilateral has diagonals that are perpendicular bisectors of each other, then it is a rhombus"**: A square (a type of rhombus) and rhombus have this property, but a kite has perpendicular diagonals (not bisectors), and a rhombus is the only one with perpendicular bisectors? Wait, no—actually, a rhombus (and square) has diagonals that are perpendicular bisectors. But is there a counterexample? Wait, no—if diagonals are perpendicular bisectors, it must be a rhombus. Wait, no, actually, a square is a rhombus. Wait, maybe I misremember. Wait, the definition of a rhombus is a quadrilateral with all sides equal, and diagonals that are perpendicular bisectors. So if diagonals are perpendicular bisectors, then it's a rhombus. But wait, is there a quadrilateral with perpendicular bisecting diagonals that's not a rhombus? No, because that's the definition. Wait, no, maybe not. Wait, let's check other options.

2. **"If a quadrilateral has all sides congruent, then it is a square"**: A rhombus has all sides congruent but is not a square (unless it's a square). So a rhombus is a counterexample (all sides congruent, not a square). So this statement has a counterexample.

3. **"If a quadrilateral has at least two angles that are congruent, then it is a trapezoid"**: Many quadrilaterals have two congruent angles (e.g., a rectangle has all angles congruent, a parallelogram has opposite angles congruent, a kite has two pairs of congruent angles, a triangle? No, quadrilateral. Wait, a rectangle is not a trapezoid (in some definitions, a trapezoid has exactly one pair of parallel sides; in others, at least one). But even so, a square (with four congruent angles) is not a trapezoid in the "exactly one" definition. Also, a parallelogram (with two pairs of congruent angles) is not a trapezoid in the "exactly one" definition. So this has counterexamples (e.g., a parallelogram has two pairs of congruent angles but is not a trapezoid in the exclusive definition, or even in the inclusive, a parallelogram is a trapezoid? Wait, definitions vary. But regardless, a rectangle (with four congruent angles) is a parallelogram, and if trapezoid is defined as at least one pair, then a rectangle is a trapezoid. But the statement is "at least two angles congruent" implies trapezoid. A kite has two pairs of congruent angles (the ones between the unequal sides? Wait, a kite has one pair of opposite angles congruent? No, a kite has two distinct pairs of adjacent sides equal, and one pair of opposite angles congruent. Wait, maybe a kite has two congruent angles. But a kite is not a trapezoid (no parallel sides). So a kite is a counterexample: it has two congruent angles but is not a trapezoid. So this statement has a counterexample.

4. **"If a quadrilateral has two pairs of congruent sides, then it is a parallelogram"**: A kite has two pairs of congruent adjacent sides (so two pairs of congruent sides) but is not a parallelogram (unless it's a rhombus, but a kite with two distinct pairs of adjacent sides is not a parallelogram). So a kite is a counterexample (two pairs of congruent sides, not a parallelogram). Wait, no—wait, a parallelogram has two pairs of congruent opposite sides. A kite has two pairs of congruent adjacent sides. So a kite has two pairs of congruent sides (adjacent) but is not a parallelogram. So this statement has a counterexample.

Wait, but the first option: "If a quadrilateral has diagonals that are perpendicular bisectors of each other, then it is a rhombus." Let's recall: In a rhombus, diagonals are perpendicular bisectors. Conversely, if diagonals are perpendicular bisectors, then the quadrilateral is a rhombus. Let's prove it: If diagonals bisect each other, it's a parallelogram (by the theorem that if diagonals bisect each other, it's a parallelogram). Then, if diagonals are perpendicular, a parallelogram with perpendicular diagonals is a rhombus (since in a parallelogram, diagonals bisect each other; if they are perpendicular, then the sides are equal: using Pythagoras, in the triangles formed by the diagonals, the sides are equal). So a parallelogram with perpendicular diagonals is a rhombus. And if diagonals are bisectors (so it's a parallelogram) and perpendicular, then it's a rhombus. So if diagonals are perpendicular bisectors, then it's a parallelogram (because bisecting) and perpendicular, so rhombus. Therefore, there is no counterexample. Wait, but earlier I thought maybe, but no—if diagonals are perpendicular bisectors, then it's a rhombus (and square, which is a rhombus). So this statement is true, no counterexample.

Wait, but let's recheck the options. The first option: "If a quadrilateral has diagonals that are perpendicular bisectors of each other, then it is a rhombus." So if diagonals are perpendicular bisectors, then it's a rhombus. Is there a quadrilateral with diagonals that are perpendicular bisectors that's not a rhombus? No, because:

- Diagonals bisect each other ⇒ parallelogram (by theorem: diagonals bisect each other ⇨ parallelogram).

- In a parallelogram, diagonals are perpendicular ⇒ rhombus (by theorem: in a parallelogram, if diagonals are perpendicular, then it's a rhombus).

Therefore, if diagonals are perpendicular bisectors, then it's a parallelogram (from bisecting) and diagonals are perpendicular (from the condition), so it's a rhombus. Therefore, this statement has no counterexample.

The other options:

- "All sides congruent ⇒ square": Rhombus is a counterexample (all sides congruent, not a square).

- "At least two angles congruent ⇒ trapezoid": Kite (two congruent angles, not a trapezoid) or parallelogram (two pairs, not a trapezoid in exclusive definition) is a counterexample.

- "Two pairs of congruent sides ⇒ parallelogram": Kite (two pairs of adjacent congruent sides, not a parallelogram) is a counterexample.

Therefore, the first statement ("If a quadrilateral has diagonals that are perpendicular bisectors of each other, then it is a rhombus") has no counterexample.

# Answer:
The first option: "If a quadrilateral has diagonals that are perpendicular bisectors of each other, then it is a rhombus" (the first radio button option)

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