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File: PythagorasByIncircle

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''' <img src="/cgi-bin/CountHits.py?PythagorasByInCircle" alt="" />
********
!! An old proof, rediscovered
Some 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.

''' <table cellpadding="5" align=right STYLE="margin-left:15px;" border=1 width="50%"><tr><td>
''' <img src="/images/PythagorasByInCircle0.png" border=0 width="100%" alt="PythagorasByInCircle0.png">
''' </td></tr></table>
So take a right-angled triangle, and inscribe an incircle.
Suppose the circle is of radius $r$, and the point of
''' contact
with the hypotenuse divides it into lengths $a$ and $b.$  Then
we have the shorter sides as $a+r$ and $b+r,$ as shown.

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

|>> $\frac{1}{2}(a+r)(b+r)~=~ar+br+r^2$ <<|

Multiply through by 2, expand and simplify:

|>> $ab+ar+br+r^2~=~2(ar+br+r^2)$ <<|

|>> $ab~=~r^2+ar+br$ <<|

So now let's ask about the square on the hypotenuse.

|>> $(a+b)^2~=~a^2+2ab+b^2$ <<|

Substituting in the value for $ab$ we get:

|>> $(a+b)^2~=~a^2+2(r^2+ar+br)+b^2$ <<|

Rearrange:

|>> $(a+b)^2~=~(a^2+2ar+r^2)~+~(b^2+2br+r^2)$ <<|

So:

|>> $(a+b)^2~=~(a+r)^2~+~(b+r)^2$ <<|

And we're done.
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