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Given P is any point in the interior of the rectangle ABCD. Show BP^2 + PD^2 = AP^2 + CP^2. Is the result the same when P in the exterior of the rectangle?

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Given P is any point in the interior of the rectangle ABCD. Show BP^2 + PD^2 = AP^2 + CP^2. Is the result the same when P in the exterior of the rectangle?

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Solution Summary

A geometry proof is analyzed. The relationship between vertices of a rectangle and another point is proven for the exterior of the rectangle.

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Let the height of the rectangle be h (=AD=BC) and the width be w (=AB=BC)

Let the vertical projection of P on AB be at distance x from A,
then it is at distance (w-x) from B

Let the horizontal projection of P on AD be at distance y from ...

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