If ABCD is a rectangle and P is any point within it, how would you prove that AP^2 + PC^2 = PB^2 + PD^2 ?

Question

Please help!! I would really appreciate it!!

Answer 1

Observe that AP^2 + PC^2 = AC^2 and PB^2 + PD^2 = DB^2, according to the Pythagorean Theorem. Observe that AC and DB are the diagonals of the rectangle. The diagonals of a rectangle have equal length. Thus, AC = DB. Observe that squaring both sides of an equation maintains equality. Hence, AC^2 = DB^2. Therefore, AP^2 + PC^2 = PB^2 + PD^2.

Answer 2

this become my first reaction: ok, per chance i'm obtuse, yet i'd say the probability that the attitude APB is sharp is countless. there is in uncomplicated phrases one consider the sq. which will produce a ninety degree perspective, that's the middle factor of the sq., and there are not any factors in the sq. which will produce an perspective more advantageous than ninety tiers. so i guess i should be obtuse. it would look that any factor P in route of AB than the middle of the sq. will produce an obtuse perspective, at the same time as any factor P farther from AB than the middle of the sq. will produce a pointy perspective. If that's real, then the probability is 50% or only below. yet is this real? upon added mirrored image: i'd say that in case you draw a semicircle from A to the middle of the sq. to B, you've defined the set of all factors which will produce a ninety degree perspective APB. any factor P interior that semicircle will produce an obtuse perspective, and any factor P outside that semicircle will produce a pointy perspective APB component to the semicircle is pi/8 and component to the sq. is a million so i guess the probability is 7pi/8 p.c., minus the factors on the semicircle itself.

Answer 3

In the first part of the ecuation you have the AB*AB, that means the hipotenuse. If that is ecqual, then CD*CD will be the same distance betwen AB*AB:
AB^2=CD^2
Can be too, AC^2=BD^2, but that distance kud be different to AB^2=CD^2. Now, if AB^2=CD^2=AC^2=BD^2, we talk about a square.
Check the Pitagoras theory....

Answer 4

pb+22

Answer 5

yeah its using Pythagoras theorem...a diagram may be helpful...use it

Answer 6

with pythagorean's theorm

Answer 7

I...I...I..dont...KNOW ABOUT THAT...can you teach me?? I am second year and if you have time please teach me... I want to