In a classroom,$4$ friends are seated at the points $A, B, C$ and $D$ as shown in the figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli,"Don't you think $ABCD$ is a square?" Chameli disagrees. Using the distance formula,find which of them is correct.

(N/A) The positions of the $4$ friends are $A(3, 4)$,$B(6, 7)$,$C(9, 4)$,and $D(6, 1)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(6 - 3)^2 + (7 - 4)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$
$BC = \sqrt{(9 - 6)^2 + (4 - 7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$
$CD = \sqrt{(6 - 9)^2 + (1 - 4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$
$DA = \sqrt{(3 - 6)^2 + (4 - 1)^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$
Now,calculate the diagonals:
$AC = \sqrt{(9 - 3)^2 + (4 - 4)^2} = \sqrt{6^2 + 0^2} = 6$
$BD = \sqrt{(6 - 6)^2 + (1 - 7)^2} = \sqrt{0^2 + (-6)^2} = 6$
Since all sides are equal $(AB = BC = CD = DA = 3\sqrt{2})$ and both diagonals are equal $(AC = BD = 6)$,the quadrilateral $ABCD$ is a square. Therefore,Champa is correct.