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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.

 Sep 10, 2017
 #1
avatar+98090 
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Look at the following image :

 

 

Let P be some random point in rectangle ABCD

 

Draw GH parallel to AB and EF parallel to AD

 

And by the Pythagorean Theorem, we have

 

PA^2  = AE^2 + EP^2   →  EP^2  = PA^2 - AE^2    (1)

PD^2 =  AE^2 + PF^2  →  PF^2  = PD^2 - AE^2    (2)

PC^2  = PF^2 + EB^2   →  PF^2  = PC^2 - EB^2   (3)

PB^2 = EB^2 + EP^2   →   EP^2  = PB^2 - EB^2    (4)

 

Equate (1), (4)  and  (2), (3)

PA^2 - AE^2 =  PB^2 - EB^2  →  PA^2 + EB^2  = PB^2 + AE^2     (5)

PD^2 - AE^2 = PC^2 - EB^2   →  PD^2 + EB^2 =  PC^2 + AE^2    (6)

 

Subtract (6) from (5)

 

PA^2 - PD^2  =  PB^2 - PC^2

 

Rearrange as

 

PA^2 + PC^2  =   PB^2 + PD^2

 

cool cool cool

 Sep 10, 2017
edited by CPhill  Sep 10, 2017

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