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Answer :
In parallelogram QPQR diagonals QO and PR intersect at S. The statement is c, A parallelogram is a simple quadrilateral with two pairs of parallel sides. We can have more arguments to prove that PQRS is a rhombus, but, the argument that we will use here is: Let's look at the first statement, we have PT> QT That's not correct, it would just prove that QR/2 > PS/2, PR = QS This statement implies PR² = QS² PS² + SR² = PQ² +QR² We cannot conclude that PS + SR = PQ =QR The next statement is PT = QT A rhombus can have different diagonals, and in fact, they have. Then let's go to the next one ST = QT It also does not exactly say it's a rhombus, it's a parallelogram property. By doing that we have that the diagonal bisects the angle That implies that angle b is also bisected. The last statement is That's the vertex angle, it's true always, not only in that case, therefore the only possible answer is Therefore, the correct option is C, Learn more about parallelograms, here: https://brainly.com/question/29274306 #SPJ1 The question is incomplete. The options are added in the picture.What is parallelogram?
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