Question

if abcd be a rectangle and p be any point in the plane of the rectangle,then prove that pa^2+pc^2=pb^2+pd^2

Answer 1

from the diagram,
PD² - DA'² = PA'² = PA² - AA'²

similarly, 
PC² - CC'² = PC'² = PB² - BC'²

hence,
PD² -  PA² = PC² - PB²
⇒PD² + PB² = PC² + PA²

Answer 2

Step-by-step explanation:

construct a line parralel to AD through P

use pythagoras theorem

pa^2=pe^2+ae^2

pb^2=be^2+pe^2

pc^2=pf^2+cf^2

pd^2=pf^2+fd^2

but AE=FD

EB=FC

so PA^2+PC^2=PB^2+PD^2

OR

construct a line parallel to AD through P and let it intersect AB at E and CD at F

use pythagoras theorem

pa^2=pe^2+ae^2

pb^2=pe^2+be^2

pc^2=pf^2+cf^2

pd^2=pf^2+df^2

AE=FD && BE=CF

So PA^2+PC^2=PE^2+PF^2+AE^2+CF^2=PE^2+PF^2+FD^2+BE^2=PB^2+Pd^2