Question 1022906
Consider the parallelogram ABCD on the Cartesian plane, with coordinates A(0,0), B(a,0), C(c+a,d), and D(c,d).  Without loss of generality, we can assume a>0 and {{{c>=0}}}.  (We wish to show that c = 0.)
The two diagonals are AC and BD.

Now apply the distance formula to the endpoints of AC and BD:

{{{abs(AC)^2 = abs(BD)^2}}}
==> {{{(c+a)^2 + d^2 = (c-a)^2+d^2}}}

==> {{{c^2 +2ac+a^2 +d^2 = c^2 -2ac+a^2 + d^2}}}, after expansion;

==> 4ac = 0, after simplification.

Since a>0, we then have c = 0. (This implies that angle DAB and angle ABC are right angles.)

Therefore parallelogram ABCD is a rectangle.