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ajinkya bhole Grade: 10
        

 



Consider a rectangle


ABCD.


 Let P be any point in the plane of this rectangle.


Prove that



PA2 + 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?

6 years ago

Answers : (2)

AskiitiansExpert Abhinav Batra
25 Points
										

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

6 years ago
AskiitiansExpert Abhinav Batra
25 Points
										

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^21793_14504_rectangle.jpg

6 years ago
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