Question 1027077
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Point P is inside rectangle ABCD. Show that
PA^2 + PC^2  = PB^2 + PD^2.

Be sure that your proof works for ANY point inside the rectangle.
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<pre>
Let us assume that 

  - the vertice A of the rectangle has the coordinates (0,0)   (lies at the origin)
  - the vertice B                                      (a,0),
  - the vertice C                                      (a,b),   and
  - the vertice D                                      (0,b).

Let the point P is (x,y).

Then 

{{{PA^2 + PC^2}}} = {{{x^2 + y^2 + (a-x)^2 + (b-y)^2}}}.

{{{PB^2 + PD^2}}} = {{{x^2 + (y-b)^2 + (a-x)^2 + y^2}}},

according to the Pythagorean theorem.

Compare these formulas to make sure that the right sides are equal.

So the left sides are.

Proved.
</pre>