Proof Question

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.

Use pythagoras theorem to find each of these 4 terms and it falls out easily.

Use this diagram

Very nice Melody.....!!!!.......here's the math.....

PA^2  = (AD -v)^2 + w^2

PC^2  = (CD - w)^2  + v^2

PB^2  = (CD -w)^2 + (AD - v)^2

PD^2  = v^2 + w^2

So

PA^2 + PC^2   = PB^2 + PD^2   ???

(AD -v)^2 + w^2 + (CD - w)^2  + v^2  = (CD -w)^2 + (AD - v)^2 + v^2 + w^2  ???

Rearrange the terms  on the right

(AD -v)^2 + w^2 + (CD - w)^2  + v^2  = (AD -v)^2 + w^2 + (CD - w)^2  + v^2

And it's true  !!!!

[P. S.  - I liked this one !!!  ]