# Pythagoras by Incircle

## 2016/06/08 …

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.

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\$

\$(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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