Flag Analytical Geometry> rectancle...
question mark

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?

ajinkya bhole , 15 Years ago
Grade 10
anser 2 Answers
AskiitiansExpert Abhinav Batra

Last Activity: 15 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

AskiitiansExpert Abhinav Batra

Last Activity: 15 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^21793_14504_rectangle.jpg

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