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```

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?```
7 years ago

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
```
7 years ago
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^2
```
7 years ago
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