Question 823783
{{{drawing(300,300,-0.9,0.9,-0.9,0.9,
grid(0),
line(0.6,0,0,0.8),line(0.6,0,0,-0.8),
line(-0.6,0,0,0.8),line(-0.6,0,0,-0.8),
locate(0.58,0.11,A(2a,0)),locate(0.03,0.85,B(0,2b)),
locate(-0.54,0,C(-2a,0)),locate(0.03,-0.77,D(0,-2b)),
red(circle(0.3,0.4,0.02)),red(circle(0.3,-0.4,0.02)),
red(circle(-0.3,0.4,0.02)),red(circle(-0.3,-0.4,0.02)),
locate(0.33,0.45,E),locate(0.33,-0.35,H),
locate(-0.38,0.45,F),locate(-0.38,-0.35,G)
)}}} E, F, G, and H are the midpoints of sides AB, BC, CD, and DA respectively.
 
Step 1: Determine the coordinates of the midpoints
{{{x[E]=(x[A]+x[B])/2=(2a+0)/2=2a/2=a}}}
{{{y[E]=(x[A]+x[B])/2=(0+2b)/2=2b/2=b}}}
and so on.
 
Step 2: Show that {{{y[E]=b=y[F]}}} , which proves that 
EF is part of the horizontal line {{{y=b}}} .
Similarly prove that {{{x[F]=-a=x[G]}}} , which proves that
FG is part of the vertical line {{{x=-a}}} .
You can similarly prove that
GH is part of the horizontal line {{{y=-b}}} , and
HE is part of the vertical line {{{x=a}}} .
 
Step 3:
State that
two horizontal lines are parallel to each other, so EF is parallel to GH,
and that
two vertical lines are parallel to each other, so FG is parallel to HE.
Then state that since opposite sides are parallel, quadrilateral EFGH is a parallelogram.
 
Step 4:
State that horizontal lines are perpendicular to vertical lines, so
EF is perpendicular to FG (meaning that angle EFG is a right angle),
FG is perpendicular to GH (meaning that angle FGH is a right angle),
GH is perpendicular to HE (meaning that angle GHE is a right angle), and
HE is perpendicular to EF (meaning that angle HEF is a right angle).
That means that in parallelogram EFGH,
the angles at F, G, H, and E are right angles.
 
Step 5:
State that since EFGH is a parallelogram with 4 right angles, it is a rectangle.