+99 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.
+129899 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
