Question

given: abcd is a parallelogram. prove: \\(\overline{ab} \cong \overline{cd}\\) and \\(\overline{bc} \cong \overline{da}\\)
diagram of parallelogram abcd with diagonal ac, and markings on ab and cd
statements:

1. abcd is a parallelogram
2. draw \\(\overline{ac}\\)
3. \\(\angle bca\\) and \\(\angle dac\\) are alt. interior angles
4. \\(\angle dca\\) and \\(\angle bac\\) are alt. interior angles
5. \\(\overline{ab} \parallel \overline{cd}\\)

reasons:

1. given
2. unique line postulate
3. def. of alt. interior angles
4. def. of alt. interior angles
5. def. of parallelogram

options on right: angles, segments, triangles, statements, reasons; asa, def. of parallelogram, reflexive property, alternate interior angles theorem, etc.
correct! assemble the next statement.

Explanation:

Step1: Recall parallelogram properties

In a parallelogram, opposite sides are parallel. Given \(ABCD\) is a parallelogram, by definition of a parallelogram, \(AB \parallel CD\) and \(AD \parallel BC\). But here, step 5's statement is \(AB \parallel CD\), and the reason should align with the definition of a parallelogram (since a parallelogram is defined as a quadrilateral with both pairs of opposite sides parallel). Wait, but looking at the given steps, step 5's reason is "def. of parallelogram" which matches the statement \(AB \parallel CD\) because the definition of a parallelogram includes opposite sides being parallel. But maybe the next step (after step 5) would involve triangle congruence? Wait, the problem is to assemble the next statement. Wait, the current steps are 1 - 5, and we need to find the next statement. Wait, maybe I misread. Wait, the goal is to prove \(AB \cong CD\) and \(BC \cong DA\). So after establishing \(AB \parallel CD\) (step 5), we can look at triangles \(ABC\) and \(CDA\). Let's correct:

Wait, step 1: \(ABCD\) is a parallelogram (given). Step 2: Draw \(AC\) (unique line postulate). Step 3: \(\angle BCA\) and \(\angle DAC\) are alt. interior angles (def. of alt. interior angles, since \(AD \parallel BC\) from parallelogram def). Step 4: \(\angle DCA\) and \(\angle BAC\) are alt. interior angles (def. of alt. interior angles, since \(AB \parallel CD\) from parallelogram def). Step 5: \(AB \parallel CD\) (def. of parallelogram). Now, next, \(AC \cong AC\) (reflexive property). Then, \(\triangle ABC \cong \triangle CDA\) by ASA (since \(\angle BCA \cong \angle DAC\), \(AC \cong AC\), \(\angle BAC \cong \angle DCA\)). Then, by CPCTC, \(AB \cong CD\) and \(BC \cong DA\). But the next statement after step 5 (which is \(AB \parallel CD\)) should be \(AC \cong AC\) (reflexive property) or looking at the triangles. Wait, the options for statements: maybe " \(AC \cong AC\) " (reflexive) or " \(\triangle ABC \cong \triangle CDA\) " (ASA). But the given options for statements: Angles, Segments, Triangles, Statements. Wait, the next statement after step 5 ( \(AB \parallel CD\) ) would be related to the common side. So the next statement is \(AC \cong AC\) (reflexive property), but as a statement, it's " \(AC \cong AC\) " with reason "reflexive property". But maybe the next step in the proof (after step 5) is to state \(AC \cong AC\) (reflexive), then use ASA for triangle congruence.

Wait, let's re - structure the proof:

1. \(ABCD\) is a parallelogram (given)
2. Draw \(AC\) (unique line postulate)
3. \(\angle BCA\) and \(\angle DAC\) are alt. interior angles (def. of alt. interior angles, because \(AD \parallel BC\) from parallelogram definition)
4. \(\angle DCA\) and \(\angle BAC\) are alt. interior angles (def. of alt. interior angles, because \(AB \parallel CD\) from parallelogram definition)
5. \(AB \parallel CD\) (def. of parallelogram)
6. \(AC \cong AC\) (reflexive property)
7. \(\triangle ABC \cong \triangle CDA\) (ASA, using \(\angle BCA \cong \angle DAC\), \(AC \cong AC\), \(\angle BAC \cong \angle DCA\))
8. \(AB \cong CD\) and \(BC \cong DA\) (CPCTC)

So the next statement after step 5 is \(AC \cong AC\) (statement: \(AC \cong AC\), reason: reflexive property). But looking at the given options for statements (Angles, Segments, Triangles, Statements) and reasons (reflexive property, alternate interior angles theorem, def. of parallelogram, ASA). So the next statement (after step 5) should be \(AC \cong AC\) (under Segments or Statements) with reason "reflexive property".

But maybe the question is to find the next statement in the proof sequence. So after step 5 (\(AB \parallel CD\) with reason def. of parallelogram), the next statement is \(AC \cong AC\) (statement) with reason reflexive property.

Answer:

The next statement is \(AC \cong AC\) (with reason "reflexive property")