How can we Prove pythagore theorem?

Answer 1

There are several different proofs on the web page below.

I remember visiting another web page with about fifty different proofs, but I can't find it again.

Answer 2

pythagoras theorem - ABsquare+ BCsquare=ACsquare
and only for right angled triangles
hehehe the simplest and the most easiest[ childish ] proof is-
let AB=3and BC= 4 and AC=5
AB square=9 and BCsquare=16
ABsq.+BCsq.=9+16=25=ACsq.
hence the theorem is proved isnt it? hehehe

Answer 3

Pythagorean Theorem

72 Proofs

Click on the URL below for additional information concerning Pythagorean Theorem Proofs

www.cut-the-knot.org/pythagoras/index.shtml

- - - - - - -s-

Answer 4

See for the proofs the link below:

Answer 5

I can't write that here completely!!!!!
Check out these links, some of them are higher level though

http://www.cut-the-knot.org/pythagoras/index.shtml

http://mathforum.org/isaac/problems/pythagthm.html

http://72.14.235.104/search?q=cache:opedoKhXd1cJ:www.pbs.org/teachersource/mathline/concepts/historyandmathematics/act1wks.pdf+prove+pythagoras+theorem&hl=en&gl=in&ct=clnk&cd=6

http://www.sunsite.ubc.ca/DigitalMathArchive/Euclid/java/html/pythagoras.html

http://www.sunsite.ubc.ca/DigitalMathArchive/Euclid/java/html/pythagoras.html

Answer 6

we can use vectors

assume we have vector AC perpendicular to vector AB
BC^2 = (vector BC)^2 = (vectorAC - vectorAB)^2
= (vectorAB)^2 + (vectorAC)^2 - 2(vectorAB)(vectorAC)
= AB^2 + AC^2 - 2 AB*AC*cos BAC
= AB^2 + AC^2 (cosBAC = 0 )

Answer 7

http://www.cut-the-knot.org/pythagoras/index.shtml

Answer 8

distance formula