Four pegs are driven in at A(0, 0), B(7, 1), C(9, 6) and D(2, 5). Opposite sides come out equal in pairs. Compute the two diagonals and decide what ABCD is.

$AB=CD=\sqrt{49+1}=\sqrt{50}$ and $BC=DA=\sqrt{4+25}=\sqrt{29}$, so ABCD is a parallelogram. Its diagonals are $AC=\sqrt{(9-0{)}^2+(6-0{)}^2}=\sqrt{117}$ and $BD=\sqrt{(2-7{)}^2+(5-1{)}^2}=\sqrt{41}$. They are unequal, so the parallelogram is not a rectangle; and since $\sqrt{50}$ is not $\sqrt{29}$, it is not a rhombus either. Equal opposite sides alone never prove a rectangle.