Pythagoras By Incircle 


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An old proof, rediscoveredSome time ago I was working on a puzzle about incircles, and unexpectedly a proof of Pythagoras' Theorem dropped out! I'm sure it's well known to people who know lots about Pythagoras' Theorem, but I thought I'd share it.
Let's think about the area of the triangle. On the one hand it's $\frac{1}{2}(a+r)(b+r),$ but on the other hand it's the square plus each kite, which is $r^2+ar+br.$ That is not instantly obvious, but is left as an exercise for the interested reader. It might help to remember that the centre of the incircle lies on the angle bisectors. Equating these two expressions for the area we get
Multiply through by 2, expand and simplify:
So now let's ask about the square on the hypotenuse.
Substituting in the value for $ab$ we get:
Rearrange:
So:
And we're done.
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Quotation from Tim BernersLee 