Question

1. ( jklm ) is a rhombus. theorem 6.3c
2. match the correct answers.

given: ( overline{ac} cong overline{bd} )
image of parallelogram ( abcd ) with diagonals ( ac ) and ( bd )
is the parallelogram a square?
write the number of the theorem that supports your answer

1. ( abcd ) is a rectangle. theorem
2. ( abcd ) is a rhombus. theorem

Explanation:

Step1: Recall properties of parallelograms, rectangles, rhombuses, and squares

A parallelogram has opposite sides equal and parallel. A rectangle is a parallelogram with congruent diagonals. A rhombus is a parallelogram with all sides congruent. A square is a parallelogram that is both a rectangle and a rhombus (i.e., has congruent diagonals and all sides congruent).

Given that \( \overline{AC} \cong \overline{BD} \) (diagonals are congruent) and \( ABCD \) is a parallelogram. First, a parallelogram with congruent diagonals is a rectangle (Theorem for rectangle: if a parallelogram has congruent diagonals, it is a rectangle). Now, we also need to check if it's a rhombus. But from the diagram, if we assume it's a parallelogram with congruent diagonals (so rectangle) and if we consider the sides, but the key here is: To be a square, a parallelogram must be both a rectangle (congruent diagonals) and a rhombus (all sides congruent). But the question first: "Is the parallelogram a square?" Wait, no, wait the given is \( \overline{AC} \cong \overline{BD} \), so first, for the first part: "Is the parallelogram a square?" Wait, maybe the diagram is a parallelogram (let's assume \( ABCD \) is a parallelogram). Wait, the first match: Given \( \overline{AC} \cong \overline{BD} \), so a parallelogram with congruent diagonals is a rectangle (Theorem 1: \( \parallelogram ABCD \) is a rectangle). Then, if a parallelogram is a rhombus (all sides equal) and a rectangle (congruent diagonals), it's a square. But the question "Is the parallelogram a square?" – wait, maybe the diagram is a rhombus? Wait, no, the given is \( \overline{AC} \cong \overline{BD} \). Wait, let's re-express:

1. For the first dropdown: "Is the parallelogram a square?" – If the parallelogram is both a rectangle (diagonals congruent) and a rhombus (all sides equal), then yes. But from the given \( \overline{AC} \cong \overline{BD} \), we know it's a rectangle. If we also know it's a rhombus (from the diagram, maybe it's a rhombus with congruent diagonals), then it's a square. But maybe the first part: the theorem for rectangle is "If a parallelogram has congruent diagonals, then it is a rectangle" (Theorem 1), and the theorem for rhombus with congruent diagonals being square: A rhombus with congruent diagonals is a square. But the question is to match:

Wait, the problem is "Match the correct answers". The first part: Given \( \overline{AC} \cong \overline{BD} \), so \( ABCD \) is a parallelogram with congruent diagonals, so it's a rectangle (Theorem 1: \( \parallelogram ABCD \) is a rectangle). Then, if a parallelogram is a rhombus (Theorem 2: \( \parallelogram ABCD \) is a rhombus) and a rectangle, it's a square. But the question "Is the parallelogram a square?" – if it's a rhombus (all sides equal) and a rectangle (congruent diagonals), then yes. But maybe the first match:

Wait, let's structure:

- Given \( \overline{AC} \cong \overline{BD} \), so \( ABCD \) (parallelogram) has congruent diagonals → so it's a rectangle (Theorem 1: \( \parallelogram ABCD \) is a rectangle).

- Then, if \( ABCD \) is a rhombus (Theorem 2: \( \parallelogram ABCD \) is a rhombus) and a rectangle, then it's a square.

But the question "Is the parallelogram a square?" – if it's both a rectangle (from diagonals) and a rhombus (from sides), then yes. But maybe the diagram is a rhombus with congruent diagonals, so it's a square.

But the main matches:

1. \( \parallelogram ABCD \) is a rectangle. Theorem: If a parallelogram has congruent diagonals, it is a rectangle. So this matches with the given \( \overline{AC} \cong \overline{BD} \).

2. \( \parallelogram ABCD \) is a rhombus. Theorem: If a parallelogram has all sides congruent, it is a rhombus (but maybe from the diagram, it's a rhombus).

But the first question: "Is the parallelogram a square?" – If it's a rhombus (Theorem 2) and a rectangle (Theorem 1), then yes. But maybe the answer for "Is the parallelogram a square?" is yes (if it's both), but let's focus on the theorem matches.

First, the given \( \overline{AC} \cong \overline{BD} \), so the parallelogram with congruent diagonals is a rectangle → so Theorem 1: \( \parallelogram ABCD \) is a rectangle.

Then, if \( ABCD \) is a rhombus (Theorem 2), and a rectangle, then it's a square.

But maybe the problem is:

- Match "Given \( \overline{AC} \cong \overline{BD} \)" with "1. \( \parallelogram ABCD \) is a rectangle. Theorem" because a parallelogram with congruent diagonals is a rectangle.

- Then, "Is the parallelogram a square?" – If it's a rhombus (Theorem 2) and a rectangle, then yes. But maybe the diagram is a square, so the answer is yes, and the theorem for square is a rhombus with congruent diagonals (or a rectangle with all sides equal).

But let's proceed step by step.

Step1: Identify the property of parallelogram with congruent diagonals

A parallelogram with congruent diagonals is a rectangle. So the theorem for "1. \( \parallelogram ABCD \) is a rectangle. Theorem" is the one that applies to \( \overline{AC} \cong \overline{BD} \).

Step2: Identify the property of square

A square is a parallelogram that is both a rectangle (congruent diagonals) and a rhombus (all sides congruent). So if \( ABCD \) is a rhombus (Theorem 2) and a rectangle (Theorem 1), then it's a square.

But the question "Is the parallelogram a square?" – if we have a parallelogram that is a rectangle (from diagonals) and a rhombus (from sides), then yes.

But the main matches:

- Given \( \overline{AC} \cong \overline{BD} \) → matches with "1. \( \parallelogram ABCD \) is a rectangle. Theorem" because congruent diagonals in a parallelogram imply it's a rectangle.

- Then, if \( ABCD \) is a rhombus (Theorem 2), then it's a square (since a rhombus with congruent diagonals is a square).

So the answers would be:

- Is the parallelogram a square? Yes (if it's both a rectangle and a rhombus).

- Theorem for rectangle: 1. \( \parallelogram ABCD \) is a rectangle. Theorem.

- Theorem for rhombus: 2. \( \parallelogram ABCD \) is a rhombus. Theorem.

But maybe the problem is to match the given \( \overline{AC} \cong \overline{BD} \) with the rectangle theorem (1), and then the square is a rhombus with congruent diagonals, but let's confirm.

Wait, the standard theorems:

- Theorem for rectangle: In a parallelogram, if the diagonals are congruent, then it is a rectangle.

- Theorem for rhombus: In a parallelogram, if all sides are congruent, then it is a rhombus.

- Theorem for square: A parallelogram is a square if and only if it is both a rectangle and a rhombus (i.e., diagonals congruent and all sides congruent).

Given \( \overline{AC} \cong \overline{BD} \) (diagonals congruent), so \( ABCD \) (parallelogram) is a rectangle (Theorem 1). If \( ABCD \) is also a rhombus (Theorem 2), then it's a square.

So the first match: Given \( \overline{AC} \cong \overline{BD} \) → matches with "1. \( \parallelogram ABCD \) is a rectangle. Theorem".

Then, "Is the parallelogram a square?" → Yes (if it's a rhombus too, which from the diagram, maybe it is, as it looks like a square's diagram with diagonals).

Then, the theorem for rhombus is "2. \( \parallelogram ABCD \) is a rhombus. Theorem".

Answer:

• Is the parallelogram a square? Yes (assuming it's a rhombus with congruent diagonals).
• Theorem for rectangle: 1. \( \parallelogram ABCD \) is a rectangle. Theorem (matches \( \overline{AC} \cong \overline{BD} \)).
• Theorem for rhombus: 2. \( \parallelogram ABCD \) is a rhombus. Theorem (if it's a rhombus, which the diagram suggests).

(Note: The exact answer depends on the diagram's implicit properties, but based on diagonals congruent, it's a rectangle, and if it's a rhombus, then a square.)