Pythagoras theorem.

There is a right-angled triangle ABC,∠C=90∘
Let AB=a,BC=b,CA=c
Construct CD┴AB such that CD and AB intersect at D.
Let AD=a1,and DB=a2
Consider right-angled triangle ACD.
∠ACD=90∘-∠CAD
Consider right-angled triangle ABC.
∠ABC ∠CAB 90∘=180∘(∠sum of △)
∠ABC=90∘-∠CAB
=90∘-∠CAD
i.e.∘∠ACD=∠ABC
∠CAD=∠BAC(common angle)
∠CDA=∠BCA=90∘
∴△ACD〜△ABC(AAA)Similarly,we can prove that △CBD〜△ABC.
Since the corresponding sides of similar triangles are proportional,wehave
AC/AB=AD/AC
i.e.b/c=c1/b
b^2=c*c1
and CB/AB=BD/BC
i.e.a/c=c2/a
a^2=c*c2
∴a^2 b^2=c*c2 c*c1
=c(c2 c1)
=c*c
=c^2

2008-04-26 17:55:28 補充：
There is a right-angled triangle ABC,∠C=90∘
The answer upper is wrong
Correct Answer:
i.e.c/a=a1/c
c^2=a*a1
and CB/AB=BD/BC
i.e.b/a=a2/b
b^2=a*a2
∴b^2+c^2=a2+ a*a1
=a(a2+ac1)
=a*a
=a^2

由於需要圖畫協助解釋，請參考以下網頁：

http://en.wikipedia.org/wiki/Pythagorean_theorem

第一個proof會比較簡單同容易理解。