Question

In parallelogram PQRS, diagonals PR and QS intersect at point T.Which statement would prove PQRS is a rhombus?PT > QTPT QTPR QSSTQT

Answer 1

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

`\begin{gathered} PR^2=QS^2 \\ \\ PS^2+SR^2=PQ^2+QR^2 \end{gathered}`

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`

That also not exactly says it's a rhombus, it's a pallelogram property.

`\angle SPT=\angle QPT`

By doing that we have that the diagonal bissects the angle

That implies that the angle b is also bissect.

The last statment is

`\angle PTQ=\angle STR`

That's literally the vertex angle, it's true always, not only in that case, therefore the only possible answer is

`\angle SPT=\angle QPT`

Pro