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# Consider a rectangleABCD. Let P be any point in the plane of this rectangle. Prove thatPA2+ PC2= PB2+ PD2(plzz can u xplain with diagram)Can u also tell me what TONCAS means and what is its significance in this problem?

11 years ago

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

11 years ago

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 