Question 397109
Let A(0,0), B(a,0), C(0,a), and D(a,a) be the vertices of the square.
Then line AD has slope {{{(a - 0)/(a - 0) = 1}}}, while
line BC has slope {{{(0-a)/(a-0) = -a/a = -1}}}.
Hence the slopes are negative reciprocals of each other, and the lines AD and BC are perpendicular.