Question 1208799
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E = intersection of diagonals AC and BD
By construction we'll have EA = EC and EB = ED be the case.


Let's say we placed point E at the origin (0,0).
Let's place point A at (p,q) where p and q are any real numbers. I'll select p = 2 and q = 3 as shown in the diagram below.
To go from E to A we go p units right and q units up.
To go from E to C, we go in reverse: go p units left and q units down. This places C at (-p,-q)
By construction, AE = EC to show that diagonal AC is bisected.
You can use the distance formula, or Pythagorean theorem, to show that {{{AE = sqrt(p^2+q^2)}}} and {{{EC = sqrt(p^2+q^2)}}}, to arrive at {{{AE = EC}}}


Now let's say point B is located at (r,s) where r and s are any real numbers. I'll pick r = 3 and s = 1.
Follow similar logic as the previous paragraph to determine point D is located at (-r,-s). 
A similar construction leads to BE = ED to show that diagonal BD is bisected.


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Key takeaway:
A = (p,q)
B = (r,s)
C = (-p,-q)
D = (-r,-s)
E = (0,0)
{{{
drawing(400,400,-5,5,-5,5,
graph(400,400,-5,5,-5,5),
circle(2,3,0.08),circle(2,3,0.1),circle(2,3,0.12),circle(2,3,0.14),circle(3,1,0.08),circle(3,1,0.1),circle(3,1,0.12),circle(3,1,0.14),circle(-2,-3,0.08),circle(-2,-3,0.1),circle(-2,-3,0.12),circle(-2,-3,0.14),circle(-3,-1,0.08),circle(-3,-1,0.1),circle(-3,-1,0.12),circle(-3,-1,0.14),circle(0,0,0.08),circle(0,0,0.1),circle(0,0,0.12),circle(0,0,0.14),
line(2,3,3,1),line(3,1,-2,-3),line(-2,-3,-3,-1),line(-3,-1,2,3),line(2,3,-2,-3),line(3,1,-3,-1),
locate(2.2,3.2,"A"),locate(3.2,1.2,"B"),locate(-1.8,-2.8-0.2,"C"),locate(-2.8-0.5,-0.8-0.5,"D"),locate(0.2,0.2-0.2,"E")
)
}}}
In the diagram above we have p=2,q=3,r=3,s=1. 
However, you can select any four real numbers you want for p,q,r,s. 


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Compute the slope of side AB
m = (y2-y1)/(x2-x1)
m = (s-q)/(r-p)
Compute the slope of side CD
m = (y2-y1)/(x2-x1)
m = (-s-(-q))/(-r-(-p))
m = (-s+q)/(-r+p)
m = (-1(s-q))/(-1(r-p))
m = (s-q)/(r-p)
Both slope results are the same expression. To avoid dividing by zero, {{{r<>p}}}


Since slopeAB = slopeCD, we conclude that AB is parallel to CD.
Similar steps are followed to prove that BC is parallel to AD.
This concludes the proof that ABCD is a parallelogram.
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