Question

is parallelogram defg a rectangle?
diagram of parallelogram defg with diagonals intersecting at c, segments dc, cf, ec, cg marked as equal, cf labeled 52
options: yes, no

Explanation:

Step1: Recall rectangle properties

A parallelogram is a rectangle if its diagonals are equal in length. In a parallelogram, diagonals bisect each other. Here, \(DEFG\) is a parallelogram, so \(DC = CF\) and \(EC = CG\) (bisecting property). But to be a rectangle, \(DF=EG\). From the diagram, \(DF\) has segments with markings, but the key is: in a parallelogram, if diagonals are equal, it's a rectangle. Wait, but the diagram shows diagonals bisect each other (marks), but do we know if they are equal? Wait, no—wait, the problem: in the parallelogram \(DEFG\), the diagonals \(DF\) and \(EG\) are bisected at \(C\), but the markings on \(DF\) (each segment \(DC\) and \(CF\) are marked, and on \(EG\) as well? Wait, no, the diagram: \(DF\) has \(DC\) and \(CF\) marked (so \(DC = CF\)), and \(EG\) has segments marked (so \(EC = CG\)). But for a rectangle, diagonals must be equal. Wait, but maybe the diagram implies? Wait, no—wait, the question: is parallelogram \(DEFG\) a rectangle? Wait, no, wait: in a parallelogram, if diagonals are equal, then it's a rectangle. But here, the diagonals are bisected, but do we have info that they are equal? Wait, no—wait, maybe I misread. Wait, the diagram: \(DF\) has length related? Wait, no, the key is: in a parallelogram, rectangle is a parallelogram with right angles, or diagonals equal. But the diagram shows that the diagonals bisect each other (marks), but there's no indication that diagonals are equal. Wait, but wait—no, wait, maybe the problem is that in the parallelogram, if the diagonals are equal, it's a rectangle. But here, the diagonals are bisected, but we don't know if they are equal. Wait, but the answer: no? Wait, no—wait, maybe I made a mistake. Wait, no, let's recall: Theorem: A parallelogram is a rectangle if and only if its diagonals are congruent (equal in length). In the given parallelogram \(DEFG\), we only know diagonals bisect each other (since it's a parallelogram), but there's no information (from the diagram's markings) that the diagonals are equal. Wait, but the diagram: \(DF\) has segments marked (so \(DC = CF\)), and \(EG\) has segments marked (so \(EC = CG\)), but no marking that \(DF = EG\). So the diagonals are not shown to be equal, so the parallelogram is not a rectangle. Wait, but wait—maybe the diagram is different? Wait, no, the options are yes or no. Wait, maybe I messed up. Wait, no: in a parallelogram, diagonals bisect each other. For it to be a rectangle, diagonals must be equal. Since we don't have info that diagonals are equal, the answer is no? Wait, no—wait, maybe the diagram shows that diagonals are equal? Wait, no, the markings: \(DF\) has two segments (each marked), \(EG\) has two segments (each marked). But unless the markings on \(DF\) and \(EG\) are the same, we can't say. But the problem is, in the diagram, is there a way to tell? Wait, maybe the key is: in the parallelogram, if the diagonals are equal, it's a rectangle. Since we don't have that, the answer is no. Wait, but maybe I'm wrong. Wait, let's check again. The question: Is parallelogram \(DEFG\) a rectangle? A parallelogram is a rectangle iff diagonals are equal. In the diagram, diagonals bisect each other (so \(DC = CF\), \(EC = CG\)), but there's no indication that \(DF = EG\). So the diagonals are not equal, so it's not a rectangle. So the answer is no.

Step2: Conclusion

Since the parallelogram \(DEFG\) (a parallelogram) does not have diagonals shown to be equal (and a parallelogram is a rectangle only if diagonals are equal), the answer is no.

Answer:

no