Some of my favorite proofs of the Pythagorean Theorem

pythagorean.tex

% PDF output: http://people.fas.harvard.edu/~klapheke/pythagorean.pdf

% Sources:
%   1. <https://en.wikipedia.org/wiki/Pythagorean_theorem>
%   2. <http://www.faculty.umb.edu/gary_zabel/Courses/Phil%20281b/Philosophy%20of%20Magic/Arcana/Neoplatonism/Pythagoras/index.shtml.html>
%   3. <https://www.cut-the-knot.org/pythagoras/index.shtml>
%   4. <https://math.stackexchange.com/questions/803/what-is-the-most-elegant-proof-of-the-pythagorean-theorem>
%   5. <https://math.stackexchange.com/questions/540415/what-is-the-simplest-proof-of-the-pythagorean-theorem-you-know?noredirect=1>
%   6. <https://math.stackexchange.com/questions/675522/whats-the-intuition-behind-pythagoras-theorem?noredirect=1>
%   7. Mathologer: <https://www.youtube.com/watch?v=p-0SOWbzUYI>
%   8. Tangram proof: <https://www.futilitycloset.com/2014/09/18/piece-work-2/>
%   9. Vi Hart's origami proof: <https://www.youtube.com/watch?v=z6lL83wl31E>
\documentclass[12pt,landscape]{article}
\usepackage[a5paper,margin=1cm]{geometry}
\pagenumbering{gobble}
\title{Geometric Proofs of the Pythagorean Theorem}
\date{}

\usepackage[no-math]{fontspec}
\setmainfont[%
	Mapping        = tex-text,
	BoldFont       = texgyrepagella-bold.otf,
	ItalicFont     = texgyrepagella-italic.otf,
	BoldItalicFont = texgyrepagella-bolditalic.otf
]{texgyrepagella-regular.otf}
\usepackage{unicode-math}
\setmathfont{texgyrepagella-math.otf}

\usepackage[xetex,pdfusetitle]{hyperref}
\usepackage{tikz}
\usetikzlibrary{calc}

% canonical colors for a/b/c
\definecolor{a}{RGB}{126,122,196} % red
\definecolor{b}{RGB}{92,62,78} % purple
\definecolor{c}{RGB}{195,69,64} % medium blue

% primary accent color
\definecolor{d}{RGB}{152,194,185} % light blue

% secondary accents
\definecolor{e}{RGB}{203,198,79} % yellow
\definecolor{f}{RGB}{200,86,174} % pink
\definecolor{g}{RGB}{115,208,95} % light green
\definecolor{h}{RGB}{189,129,77} % brown
\definecolor{i}{RGB}{80,105,58} % dark green

\setlength\parindent{0pt}

\newcommand\sidea{\textcolor{a}{a}}
\newcommand\sideb{\textcolor{b}{b}}
\newcommand\sidec{\textcolor{c}{c}}

% \arc{center x}{center y}{radius}{start angle}{end angle}
\newcommand\arc[5]{\draw ([shift=(#4:#3)]#1,#2) arc (#4:#5:#3)}

\newcommand\ninetydeg[3][0]{\draw[thick,color=black,rotate around={#1:(#2,#3)}]($ (#2,#3+0.25) $) -- ++(0.25,0) -- ++(0,-0.25);}

\newcommand\threefourfive[5]{%
	\begin{scope}[rotate=#1,shift={#2}]
		% \draw[ultra thick,color=d,fill=d!10] (0,3) -- (4,0) -- (0,0);
		\draw[ultra thick,color=a] (0,0) -- (0,3) node[midway,#3] {$\sidea$};
		\draw[ultra thick,color=b] (0,0) -- (4,0) node[midway,#4] {$\sideb$};
		\draw[ultra thick,color=c] (0,3) -- (4,0) node[midway,#5] {$\sidec$};
		\ninetydeg{0}{0}
	\end{scope}
}
\newcommand\tffi[1]{\threefourfive{0}{#1}{right}{above}{below}}
\newcommand\tffii[1]{\threefourfive{90}{#1}{above}{left}{below}}
\newcommand\tffiii[1]{\threefourfive{180}{#1}{left}{below}{above}}
\newcommand\tffiv[1]{\threefourfive{270}{#1}{below}{right}{above}}

\newcommand\oneonesqrttwo[5]{%
	\begin{scope}[rotate=#1,shift={#2}]
		% \draw[ultra thick,color=d,fill=d!10] (0,3) -- (4,0) -- (0,0);
		\draw[ultra thick,color=a] (0,0) -- (0,2) node[midway,#3] {$\sidea$};
		\draw[ultra thick,color=b] (0,0) -- (2,0) node[midway,#4] {$\sideb$};
		\draw[ultra thick,color=c] (0,2) -- (2,0) node[midway,#5] {$\sidec$};
		\ninetydeg{0}{0}
	\end{scope}
}
\newcommand\oot[1]{\oneonesqrttwo{0}{#1}{right}{above}{below}}

\begin{document}
\begin{center} {\huge Geometric Proofs of the Pythagorean Theorem} \vfill
	\begin{tikzpicture}[baseline]
		\draw[ultra thick,color=a,fill=a!10] (0,3) -- (-3,3) -- (-3,0) -- (0,0) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (4,0) -- (4,-4) -- (0,-4) -- (0,0) -- cycle;
		\draw[ultra thick,color=c,fill=c!10] (4,0) -- (7,4) -- (3,7) -- (0,3) -- cycle;
		\tffi{(0,0)}
	\end{tikzpicture}
\end{center}

\pagebreak\section{Pythagorean proof}

\begin{tikzpicture}[baseline]
	\draw[ultra thick,color=c,fill=c!10] (4,0) -- (7,4) -- (3,7) -- (0,3);
	\tffi{(0,0)} \tffii{(0,-7)} \tffiii{(-7,-7)} \tffiv{(-7,0)}
\end{tikzpicture} \qquad
\begin{tikzpicture}[baseline]
	\draw[ultra thick,color=a,fill=a!10] (0,0) rectangle (3,3);
	\draw[ultra thick,color=b,fill=b!10] (3,3) rectangle (7,7);
	\tffi{(3,0)} \tffii{(3,-3)} \tffiii{(-7,-3)} \tffiv{(-7,0)}
\end{tikzpicture}

\par\bigskip
Arrange four similar right triangles in a square with side length $a+b$, so that the area not covered by the triangles is of area $c^2$. Rearrange the triangles so that the area uncovered is of area $a^2+b^2$.

\begin{itemize}
	\item Discovered independently by Bhāksara II.
	\item Also discovered in Zhōu dynasty China and recorded in the \textit{Zhōubìsuàn Jīng}.
\end{itemize}

Alternatively, note that the area of the left square is $(a+b)^2$, and it is covered by tiles of area $4\cdot\frac{1}{2}ab + c^2$.
Thus, $a^2 + 2ab + b^2 = 2ab + c^2$, and $a^2 + b^2 = c^2$.

\pagebreak\section{Pythagorean proof (variant)}

\begin{tikzpicture}[baseline]
	\tffiii{(-1,0)}
	\tffii{(0,0)}
	\tffi{(0,1)}
	\tffiv{(-1,1)}
	\begin{scope}[rotate=180,shift={(-1,0)}]
		\draw[ultra thick,color=a] (0,0) -- (0,3);
	\end{scope}
\end{tikzpicture}

\par\bigskip
The area of the square is $c^2$. It is covered by four triangles of area $\frac{1}{2}ab$, and a square of area $(b-a)^2 = b^2 - 2ab + a^2$.
Thus, $c^2 = 2ab + b^2 - 2ab + a^2 = a^2 + b^2$.

\pagebreak\section{Alandete's proof}

\begin{tikzpicture}[baseline,scale=0.5]
		\draw[ultra thick,color=b,fill=b!10] (-3,3) rectangle (1,7) node[pos=.5] {$b^2$};
		\draw[ultra thick,color=b,fill=b!10] (3,3) rectangle (7,-1) node[pos=.5] {$b^2$};
		\draw[ultra thick,color=b,fill=b!10] (3,-3) rectangle (-1,-7) node[pos=.5] {$b^2$};
		\draw[ultra thick,color=b,fill=b!10] (-3,-3) rectangle (-7,1) node[pos=.5] {$b^2$};
		\draw[ultra thick,color=a,fill=a!10] (-3,-3) rectangle (3,3) node[pos=.5] {$4a^2$};
		\draw[thick,color=e] (1,7) -- (7,-1) -- (-1,-7) -- (-7,1) -- cycle;
\end{tikzpicture}
\qquad
\begin{tikzpicture}[baseline,scale=0.5]
		\draw[ultra thick,color=b,fill=b!10] (-3,4)  -- (-3,3) -- (1,3)   -- (1,7) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (4,3)   -- (3,3) -- (3,-1)  -- (7,-1) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (3,-4)  -- (3,-3) -- (-1,-3) -- (-1,-7) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (-4,-3) -- (-3,-3) -- (-3,1)  -- (-7,1) -- cycle;

		\draw[ultra thick,color=b,fill=b!10] (-3,4)   -- (-3,1)  -- (-7,1) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (4,3)    -- (1,3)   -- (1,7) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (3,-4)  -- (3,-1)  -- (7,-1) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (-4,-3)  -- (-1,-3) -- (-1,-7) -- cycle;

		\draw[ultra thick,color=a,fill=a!10] (-3,-3) rectangle (3,3) node[pos=.5,color=c] {$4c^2$};
		\draw[ultra thick,color=c] (1,7) -- (7,-1) -- (-1,-7) -- (-7,1) -- cycle;
\end{tikzpicture}
% \begin{tikzpicture}[baseline,scale=0.5]
% 		\draw[ultra thick,color=c,fill=c!10] (1,7) -- (7,-1) -- (-1,-7) -- (-7,1) -- cycle;
% 		\draw[thick] (-3,4)  -- (-3,3);
% 		\draw[thick] (1,3)   -- (1,7);
% 		\draw[thick] (4,3)   -- (3,3);
% 		\draw[thick] (3,-1)  -- (7,-1);
% 		\draw[thick] (3,-4)  -- (3,-3);
% 		\draw[thick] (-1,-3) -- (-1,-7);
% 		\draw[thick] (-4,-3) -- (-3,-3);
% 		\draw[thick] (-3,1)  -- (-7,1);
% 		\draw[thick] (-3,-3) rectangle (3,3);
% \end{tikzpicture}

\par\bigskip
Arrange four squares of side length $b$ around a square of side length $2a$. The length of each convex side is $(b+2a)-b=2a$.
Draw a square at the corners of the former squares. This forms four triangles of side lengths $a$, $b$, and $c$, and four triangle-shaped gaps. Move the triangles into the gaps to form a square of side length $2c$.
Thus, $(2a)^2 + 4\cdot b^2 = (2c)^2$, and $a^2+b^2=c^2$.

\begin{itemize}
	\item Discovered by Edgardo Alandete.
\end{itemize}

\pagebreak\section{Euclid's proof}

% 1. \begin{tikzpicture}[baseline]
% 	\begin{scope}[rotate=216.87,shift={(0,-3)},scale=0.5]
% 		\draw[ultra thick,color=a,fill=a!10] (0,3) -- (-3,3) -- (-3,0) -- (0,0) -- cycle;
% 		\draw[ultra thick,color=b,fill=b!10] (4,0) -- (4,-4) -- (0,-4) -- (0,0) -- cycle;
% 		\draw[ultra thick,color=c,fill=c!10] (4,0) -- (7,4) -- (3,7) -- (0,3) -- cycle;
% 		\tffi{(0,0)}
% 		\draw[thick,color=black] (0,0) -- ++(53.13:7.4);
% 	\end{scope}
% \end{tikzpicture}
1. \begin{tikzpicture}[baseline]
	\begin{scope}[rotate=216.87,shift={(0,-3)},scale=0.4]
		\draw[ultra thick,color=a,fill=a!10] (0,3) -- (-3,3) -- (-3,0) -- (0,0) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (4,0) -- (4,-4) -- (0,-4) -- (0,0) -- cycle;
		\draw[ultra thick,color=c,fill=c!10] (4,0) -- (7,4) -- (3,7) -- (0,3) -- cycle;
		% \draw[ultra thick,color=a,fill=a!10] (0,3) -- (3,7) -- ++(-37.13:3.2) -- ++(233.13:5) -- cycle;
		% \draw[ultra thick,color=b,fill=b!10] (4,0) -- (7,4) -- ++(143.13:3.2) -- ++(233.13:5) -- cycle;
		\tffi{(0,0)}
		\arc{4}{0}{0.75}{180}{143.13} node[midway,right,yshift=3] {$\alpha$};
		\draw[thick,color=e] (0,3) -- (4,-4) -- (4,0) -- cycle;
		\draw[thick,color=e] (0,0) -- (7,4) -- (4,0) -- cycle;
	\end{scope}
\end{tikzpicture}
2. \begin{tikzpicture}[baseline]
	\begin{scope}[rotate=216.87,shift={(0,-3)},scale=0.4]
		\draw[ultra thick,color=a,fill=a!10] (0,3) -- (-3,3) -- (-3,0) -- (0,0) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (4,0) -- (4,-4) -- (0,-4) -- (0,0) -- cycle;
		\draw[ultra thick,color=c,fill=c!10] (4,0) -- (7,4) -- (3,7) -- (0,3) -- cycle;
		% \draw[ultra thick,color=a,fill=a!10] (0,3) -- (3,7) -- ++(-37.13:3.2) -- ++(233.13:5) -- cycle;
		% \draw[ultra thick,color=b,fill=b!10] (4,0) -- (7,4) -- ++(143.13:3.2) -- ++(233.13:5) -- cycle;
		\ninetydeg{0}{0}
		\draw[thick,color=e] (0,0) -- ++(53.13:7.4);
		% \draw[thick,color=black] (0,3) -- (4,-4);
		\draw[thick,color=black] (0,0) -- (7,4) -- (4,0) -- cycle;
	\end{scope}
\end{tikzpicture}
3. \begin{tikzpicture}[baseline]
	\begin{scope}[rotate=216.87,shift={(0,-3)},scale=0.4]
		\draw[ultra thick,color=a,fill=a!10] (0,3) -- (-3,3) -- (-3,0) -- (0,0) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (4,0) -- (4,-4) -- (0,-4) -- (0,0) -- cycle;
		\draw[ultra thick,color=c,fill=c!10] (4,0) -- (7,4) -- (3,7) -- (0,3) -- cycle;
		% \draw[ultra thick,color=a,fill=a!10] (0,3) -- (3,7) -- ++(-37.13:3.2) -- ++(233.13:5) -- cycle;
		% \draw[ultra thick,color=b,fill=b!10] (4,0) -- (7,4) -- ++(143.13:3.2) -- ++(233.13:5) -- cycle;
		\ninetydeg{0}{0}
		\arc{0}{3}{0.5}{270}{323.13} node[midway,left,yshift=3] {$\beta$};
		\draw[thick,color=black] (0,0) -- ++(53.13:7.4);
		\draw[thick,color=e] (0,3) -- (-3,3) -- (4,0) -- cycle;
		\draw[thick,color=e] (0,0) -- (3,7) -- (0,3) -- cycle;
	\end{scope}
\end{tikzpicture}
4. \begin{tikzpicture}[baseline]
	\begin{scope}[rotate=216.87,shift={(0,-3)},scale=0.4]
		\draw[ultra thick,color=a,fill=a!10] (0,3) -- (-3,3) -- (-3,0) -- (0,0) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (4,0) -- (4,-4) -- (0,-4) -- (0,0) -- cycle;
		% \draw[ultra thick,color=c,fill=c!10] (4,0) -- (7,4) -- (3,7) -- (0,3) -- cycle;
		\draw[ultra thick,color=a,fill=a!10] (0,3) -- (3,7) -- ++(-37.13:3.2) -- ++(233.13:5) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (4,0) -- (7,4) -- ++(143.13:3.2) -- ++(233.13:5) -- cycle;
		\ninetydeg{0}{0}
		% \draw[thick,color=e] (0,0) -- ++(53.13:7.4);
		% \draw[thick,color=black] (0,3) -- (4,-4);
		% \draw[thick,color=black] (0,0) -- (7,4);
	\end{scope}
\end{tikzpicture}

\par\bigskip
Draw two lines to form two congruent obtuse triangles with sides $b$ and $c$ and interior angle $\alpha + 90^\circ$ (1).
The area of this triangle is $\frac{1}{2}b^2$, since its base and height are $b$.
Drop a vertical line from the right angle, dividing the square with area $c^2$ into left and right rectangles (2).
The left rectangle has twice the area of the triangle, since the base of the triangle is $c$ and it's height is the width of the rectangle.
Thus, the left rectangle has area $b^2$.
By the same reasoning, the right rectangle has twice the area of the triangle with area $\frac{1}{2}a^2$ (3).
Thus, $c^2=a^2+b^2$ (4).

\pagebreak\section{Proof by bisection}

\begin{tikzpicture}[baseline]
	\begin{scope}[rotate=216.87,shift={(0,-3)}]
		\tffi{(0,0)}
		\draw[thick,dotted,color=black] (0,0) -- ++(53.13:2.4);
	\end{scope}
	% \draw[ultra thick,color=a] (0,2.4) -- (1.8,0)  node[midway,left]  {$\sidea$};
	% \draw[ultra thick,color=b] (-3.2,0) -- (0,2.4) node[midway,right] {$\sideb$};
	% \draw[ultra thick,color=c] (1.8,0) -- (-3.2,0) node[midway,above] {$\sidec$};
	% \draw[thick,dotted,color=black] (0,2.4) -- (0,0);
\end{tikzpicture}
\qquad
\begin{tikzpicture}[baseline]
	\draw[ultra thick,color=c,fill=c!10] (0,2.4) -- (1.8,0) -- (-3.2,0) -- cycle node[at={(0,1)}] {$\textcolor{c}{C}$};
	\begin{scope}[shift={(-0.2,0.2)}]
		\draw[ultra thick,color=b,fill=b!10] (-3.2,0) -- (0,0) -- (0,2.4) -- cycle node[at={(-0.75,1)}] {$\textcolor{b}{B}$};
	\end{scope}
	\begin{scope}[shift={(0.2,0.2)}]
		\draw[ultra thick,color=a,fill=a!10] (0,0) -- (1.8,0) -- (0,2.4) -- cycle node[at={(0.5,1)}] {$\textcolor{a}{A}$};
	\end{scope}
\end{tikzpicture}

\par\bigskip
Bisect the triangle at its right angle perpendicular to the hypotenuse. The two right triangles formed are similar to the original. Thus, the squares of their hypotenuses are proportional to their areas, respectively. Since sum of the areas of triangles $A$ and $B$ equals the area of triangle $C$, the sum of the squares of their hypotenuses must also be equal: $a^2+b^2=c^2$.

\begin{itemize}
	\item Discovered independently by Einstein as a schoolboy.
\end{itemize}

\pagebreak\section{Proof by rectangle construction}

% 36.87, 53.13

\begin{tikzpicture}[baseline]
	\draw[ultra thick,color=d] (0,3) -- node[midway,above] {$\sideb$}
	                           (4,3) -- node[midway,right] {$\sidea$}
	                           (4,0) -- node[midway,below right] {$\frac{\sidea\sideb}{\sidec}$}
	                           ++(233.13:2.4) -- node[midway,below left] {$\frac{\sidea\sidea}{\sidec}$}
	                           ++(143.13:3.2) -- node[midway,below left] {$\frac{\sideb\sideb}{\sidec}$}
	                           ++(143.13:1.8) -- node[midway,above left] {$\frac{\sidea\sideb}{\sidec}$}
	                           cycle;
	\tffi{(0,0)}
\end{tikzpicture}

\par\bigskip
Draw a congruent triangle whose hypotenuse is $c$. Draw two triangles similar to this, whose hypotenuses are $a$ and $b$. They form a rectangle whose long side is $c$, and $\frac{a^2}{c} + \frac{b^2}{c}$.

\pagebreak\section{Proof by scaling}

\begin{tikzpicture}[baseline]
	\begin{scope}[rotate=270,shift={(0,4.575)},scale={1.5},yscale=-1]
		\draw[ultra thick,color=a] (0,0) -- (0,3) node[midway,below] {$\sidea\sidea$};
		\draw[ultra thick,color=b] (0,0) -- (4,0) node[midway,left]  {$\sidea\sideb$};
		\draw[ultra thick,color=c] (0,3) -- (4,0) node[midway,right] {$\sidea\sidec$};
		\begin{scope}[scale=0.6666]
			\ninetydeg{0}{0}
		\end{scope}
	\end{scope}
	\begin{scope}[rotate=180,shift={(8.1,0)},scale={2},xscale=-1]
		\draw[ultra thick,color=a] (0,0) -- (0,3) node[midway,right] {$\sideb\sidea$};
		\draw[ultra thick,color=b] (0,0) -- (4,0) node[midway,below] {$\sideb\sideb$};
		\draw[ultra thick,color=c] (0,3) -- (4,0) node[midway,above] {$\sideb\sidec$};
		\begin{scope}[scale=0.5]
			\ninetydeg{0}{0}
		\end{scope}
	\end{scope}
	\begin{scope}[rotate=216.87,scale={2.5}]
		\draw[ultra thick,color=a] (0,0) -- (0,3) node[midway,left]  {$\sidec\sidea$};
		\draw[ultra thick,color=b] (0,0) -- (4,0) node[midway,below] {$\sidec\sideb$};
		\draw[ultra thick,color=c] (0,3) -- (4,0) node[midway,above] {$\sidec\sidec$};
		\begin{scope}[scale=0.4]
			\ninetydeg{0}{0}
		\end{scope}
	\end{scope}
\end{tikzpicture}

\par\bigskip
Take three right triangles and scale one by $a$, one by $b$, and one by $c$. They can be arranged into a rectangle so that sides $ac$ and $bc$ correspond. Then one side of the rectangle is of length $aa+bb$ and its opposite is of length $cc$. So $a^2+b^2=c^2$.

\pagebreak\section{Thābit ibn Qurrah's proof}

\begin{tikzpicture}[baseline,scale=0.75]
	\begin{scope}[rotate=216.87,shift={(0,-3)}]
		\draw[ultra thick,color=a,fill=a!10] (0,3) -- (-3,3) -- (-3,0) -- (0,0) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (4,0) -- (4,-4) -- (0,-4) -- (0,0) -- cycle;
		\draw[ultra thick,color=c] (4,0) -- (1,-4) -- (-3,-1) -- (0,3) -- cycle;
		\tffi{(0,0)}
		% \tffii{(-4,0)}
		% \tffiv{(0,-3)}
		\begin{scope}[rotate=90,shift={(-4,0)}]
			\draw[ultra thick,color=a] (0,0) -- (0,3) node[midway,below] {$\sidea$};
			\draw[ultra thick,color=b] (0,0) -- (4,0) node[midway,above] {$\sideb$};
			\draw[ultra thick,color=c] (0,3) -- (4,0) node[midway,above left] {$\sidec$};
			\ninetydeg{0}{0}
		\end{scope}
		\begin{scope}[rotate=270,shift={(0,-3)}]
			\draw[ultra thick,color=a] (0,0) -- (0,3) node[midway,above] {$\sidea$};
			\draw[ultra thick,color=b] (0,0) -- (4,0) node[midway,below] {$\sideb$};
			\draw[ultra thick,color=c] (0,3) -- (4,0);
			\ninetydeg{0}{0}
		\end{scope}
	\end{scope}
\end{tikzpicture}
\begin{tikzpicture}[baseline,scale=0.75]
	\begin{scope}[rotate=216.87,shift={(0,-3)}]
		\draw[ultra thick,color=c,fill=c!10] (4,0) -- (1,-4) -- (-3,-1) -- (0,3) -- cycle;
		\tffi{(0,0)}
		\begin{scope}[rotate=90,shift={(-4,-4)}]
			\draw[ultra thick,color=a] (0,0) -- (0,3) node[midway,below] {$\sidea$};
			\draw[ultra thick,color=b] (0,0) -- (4,0) node[midway,above] {$\sideb$};
			\draw[ultra thick,color=c] (0,3) -- (4,0) node[midway,above left] {$\sidec$};
			\ninetydeg{0}{0}
		\end{scope}
		\begin{scope}[rotate=0,shift={(-3,-4)}]
			\draw[ultra thick,color=a] (0,0) -- (0,3) node[midway,below] {$\sidea$};
			\draw[ultra thick,color=b] (0,0) -- (4,0) node[midway,below] {$\sideb$};
			\draw[ultra thick,color=c] (0,3) -- (4,0) node[midway,above] {$\sidec$};
			\ninetydeg{0}{0}
		\end{scope}
		\begin{scope}[rotate=270,shift={(-3,-3)}]
			\draw[ultra thick,color=a] (0,0) -- (0,3) node[midway,above] {$\sidea$};
			\draw[ultra thick,color=b] (0,0) -- (4,0) node[midway,below] {$\sideb$};
			\draw[ultra thick,color=c] (0,3) -- (4,0) node[midway,right] {$\sidec$};
			\ninetydeg{0}{0}
		\end{scope}
		\draw[thick,->] (0,0) -- (0,1);
	\end{scope}
\end{tikzpicture}

\par\bigskip
Draw a figure consisting of two squares on sides $a$ and $b$, and between them two right triangles, each congruent to the original. The area of the figure minus the three triangles is $a^2+b^2$. Draw a square on side $c$, inscribed in the figure. The parts of the figure outside this square are triangles congruent to the original, since they have the same angles and hypotenuse $c$. The area of the figure minus these three triangles is $c^2$.

% \pagebreak\section{Abū al-Wafāʾ's proof}
%
% \begin{tikzpicture}[baseline]
% 	\draw[ultra thick,color=d,fill=d!10] (4,0) rectangle (1,3);
% 	\draw[ultra thick,color=d,fill=d!10] (0,4) rectangle (1,3);
% 	\tffi{(0,0)}
% 	% \tffiii{(-4,3)}
% 	% \tffii{(7,3)}
% 	\begin{scope}[xscale=-1]
% 		\tffiv{(-4,-4)}
% 	\end{scope}
% \end{tikzpicture}
%
% \par\bigskip
% Draw a congruent triangle whose hypotenuse is at a right angle to the original triangle's hypotenuse. Draw two squares of side length $a$ and $b-a$, respectively.

\pagebreak\section{Perigal's proof}

\begin{tikzpicture}[baseline,scale=0.75]
	\draw[ultra thick,color=a,fill=a!10] (0,3) rectangle (-3,0);
	\draw[ultra thick,color=b,fill=b!10] (4,0) rectangle (0,-4);
	\draw[ultra thick,color=b] (3.5,0) -- (0.5,-4);
	\draw[ultra thick,color=b] (0,-0.5) -- (4,-3.5);
	\draw[ultra thick,color=c,fill=c!10] (4,0) -- (7,4) -- (3,7) -- (0,3) -- cycle;
	\tffi{(0,0)}
	\draw[ultra thick,color=c] (2,2) rectangle (5,5);
	\draw[ultra thick,color=c] (2,2) -- (2,1.5);
	\draw[ultra thick,color=c] (2,5) -- (1.5,5);
	\draw[ultra thick,color=c] (5,5) -- (5,5.5);
	\draw[ultra thick,color=c] (5,2) -- (5.5,2);
\end{tikzpicture}

\par\bigskip
\begin{itemize}
	\item Discovered in 1872 by Henry Perigal, and engraved on his tombstone in Essex.
\end{itemize}

\pagebreak\section{Da Vinci's proof}

\begin{tikzpicture}[baseline,scale=0.5]
	\begin{scope}[rotate=216.87]
		\draw[ultra thick,color=a,fill=a!10] (0,3) -- (-3,3) -- (-3,0) -- (0,0) -- cycle;
		\draw[ultra thick,color=b,fill=b!10] (4,0) -- (4,-4) -- (0,-4) -- (0,0) -- cycle;
		\draw[ultra thick,color=c,fill=c!10] (4,0) -- (7,4) -- (3,7) -- (0,3) -- cycle;
		\tffi{(0,0)}
		\tffiii{(-7,-7)}
		\begin{scope}[yscale=-1]
			\tffii{(0,0)}
		\end{scope}
		\draw[thick,color=e] (0,0) -- (7,7);
		\draw[thick,color=e] (-3,3) -- (4,-4);
	\end{scope}
\end{tikzpicture}

\par\bigskip
Draw a square on the hypotenuse, and a similar right triangle on the opposite side of the square, forming a hexagon. Draw two squares on the legs, and a line between them, forming another hexagon with the original triangle. Divide each hexagon along the main diagonal into two quadrilaterals. Since each quadrilateral has side lengths $b$, $c$, $a$, and identical angles, they are all congruent. Thus, the two hexagons have equal area. The area of the first, minus the two triangles, is $c^2$. The area of the second, minus the two triangles, is $a^2+b^2$.

\pagebreak\section{Garfield's proof}

\begin{tikzpicture}[baseline]
	\draw[ultra thick,color=d,fill=d!10] (0,3) -- (3,7) -- (4,0);
	\tffi{(0,0)} \tffiv{(-7,0)}
	\ninetydeg[-36.87]{0}{3}
\end{tikzpicture}

\par\bigskip
The area of the trapezoid is $(a+b)\cdot\frac{1}{2}(a+b)$, and the sum of the areas of the three triangles is $2(\frac{1}{2}ab) + \frac{1}{2}c^2$. Thus $\frac{1}{2}(a+b)^2 = ab+\frac{1}{2}c^2$; so $(a+b)^2 - 2ab = c^2$, and $a^2 + b^2 = c^2$.

\begin{itemize}
	\item Discovered by U.S.\ President James Garfield in 1876.
\end{itemize}

\pagebreak\section{Proof by circumscription}

\begin{tikzpicture}[baseline]
	\draw[ultra thick] (-5,0) arc [start angle=180,end angle=0,radius=5];
	\draw[ultra thick] (0,0) -- (5,0);
	\draw[ultra thick,color=d,fill=d!10] (5,0) -- (3,4) -- (-5,0);
	\draw[ultra thick,color=c] (-5,0) -- (0,0) node[midway,above] {$\sidec$};
	\begin{scope}
		\ninetydeg[203.58]{3}{4};
	\end{scope}
	\tffii{(0,-3)}
\end{tikzpicture}

\par\bigskip
Inscribe the triangle in a circle of radius $c$. This defines another right triangle whose hypotenuse is the diameter of the circle, and whose height $b$ splits the diameter into $c+a$ and $c-a$. Then by the intersecting chords theorem, $b\cdot b = (c+a)(c-a)$, and $b^2 = c^2 - a^2$.

\pagebreak\section{Proof by tangency}

\begin{tikzpicture}[baseline]
	\draw[ultra thick] (0,3) circle [radius=3];
	\tffi{(0,0)}
	\draw[ultra thick,color=a] (0,3) -- ++(143.13:3) node[midway,above right] {$\sidea$};
\end{tikzpicture}

\par\bigskip
Draw a circle of radius $a$ around a corner of the triangle, so that side $b$ is tangent to it.
Applying the power of a point theorem to the other non-right-angled corner, $(c+a)(c-a) = b\cdot b$, and $c^2-a^2=b^2$.

\pagebreak\section{Proof by tesselation}

\begin{tikzpicture}[baseline,scale=0.9]
	% \clip (7.2,-1.2) rectangle (16.8,-8.4);
	\foreach \x in {0,1,2,3}
		\foreach \y in {0,1,2,3}
			{
				\draw[ultra thick,color=a,fill=a!10,xshift={80*\x+34.29*(\y-\x)},yshift={-11.43*\x-45.71*(\y-\x)}] (0,0.4) -- (0,1.6) -- (1.2,1.6) -- (1.2,0.4) -- cycle;
				\draw[ultra thick,color=b,fill=b!10,xshift={80*\x+34.29*(\y-\x)},yshift={-11.43*\x-45.71*(\y-\x)}] (1.2,0) -- (1.2,1.6) -- (2.8,1.6) -- (2.8,0) -- cycle;
			}
	\draw[step=2,ultra thick,color=c,fill=c!10,rotate=36.87,shift={(0.95,1.28)}] (1,-1) grid (7,-7);
	% \draw[ultra thick,color=e,fill=c!10] (5.6,0.8) -- (7.2,2);
	\begin{scope}[xscale=-0.4,yscale=0.4]
		\threefourfive{0}{(-18,2)}{below right}{below}{above left}
	\end{scope}
\end{tikzpicture}

\par\bigskip
Tessellate the plane with squares of side length $a$ and $b$. Where these sides meet in the larger square, draw a hypotenuse $c$. Superimpose a grid of squares of side length $c$ on the plane. Each grid cell of area $c^2$ contains one square of area $a^2$ and one of area $b^2$.

% \url{http://www.cut-the-knot.org/pythagoras/PythLattice.shtml}
% \url{https://math.stackexchange.com/a/540843}

\pagebreak\section{Proof by hinged dissection}

1. \begin{tikzpicture}[baseline,scale=0.75]
	\draw[ultra thick,color=a,fill=a!10] (0,0) rectangle (3,3);
	\draw[ultra thick,color=b,fill=b!10] (3,0) rectangle (7,4);
	% \begin{scope}[rotate=0,shift={(0,0)}]
	% 	\draw[line width=0,fill=white] (0,0) -- (0,3) -- (4,0) -- cycle;
	% \end{scope}
	% \begin{scope}[rotate=90,shift={(0,-7)}]
	% 	\draw[line width=0,fill=white] (0,0) -- (0,3) -- (4,0) -- cycle;
	% \end{scope}
	% \tffi{(0,0)}
	% \tffii{(0,-7)}
\end{tikzpicture}
\quad
2. \begin{tikzpicture}[baseline,scale=0.75]
	\draw[ultra thick] (0,0) -- (0,3) -- (3,3) -- (3,4) -- (7,4) -- (7,0) -- cycle;
	% \begin{scope}[rotate=0,shift={(0,0)}]
	% 	\draw[line width=0,fill=white] (0,0) -- (0,3) -- (4,0) -- cycle;
	% \end{scope}
	% \begin{scope}[rotate=90,shift={(0,-7)}]
	% 	\draw[line width=0,fill=white] (0,0) -- (0,3) -- (4,0) -- cycle;
	% \end{scope}
	\tffi{(0,0)}
	\tffii{(0,-7)}
\end{tikzpicture}
\quad
3. \begin{tikzpicture}[baseline,scale=0.75]
	\draw[ultra thick,color=c,fill=c!10] (0,3) -- (4,0) -- (7,4) -- (3,7) -- cycle;
	\tffii{(3,-3)}
	\tffi{(3,4)}
\end{tikzpicture}

\par\bigskip
Draw two squares of area $a^2$ and $b^2$ (1). Draw two triangles inside them along the edges (2). Rearrange the three pieces to form a square of area $c^2$ (3).
\begin{itemize}
	\item Discovered by Thābit ibn Qurra in the 9th century.
\end{itemize}

\pagebreak\section{Tangram proof}

\begin{tikzpicture}[baseline]
	\begin{scope}[scale={1.5}]
		\draw[ultra thick,color=a,fill=a!10] (0,0) -- (-2,0) -- (0,2);
		\draw[ultra thick,color=a,fill=a!10] (-2,0) -- (-2,2) -- (0,2) -- (-2,0);

		\draw[ultra thick,color=b,fill=b!10] (0,0) -- (1,-1) -- (2,0);
		\draw[ultra thick,color=b,fill=b!10] (0,0) -- (0,-1) -- (1,-1) -- (0,0);
		\draw[ultra thick,color=b,fill=b!10] (0,-1) rectangle (1,-2);
		\draw[ultra thick,color=b,fill=b!10] (1,-2) -- (2,-2) -- (2,-1);
		\draw[ultra thick,color=b,fill=b!10] (1,-2) -- (2,-1) -- (2,0) -- (1,-1) -- (1,-2);

		\draw[ultra thick,color=c,fill=c!10] (2,0) -- (4,2) -- (2,2);
		\draw[ultra thick,color=c,fill=c!10] (4,2) -- (2,4) -- (2,2) -- (4,2);
		\draw[ultra thick,color=c,fill=c!10] (2,0) -- (2,1) -- (1,1);
		\draw[ultra thick,color=c,fill=c!10] (0,2) -- (1,3) -- (1,1);
		\draw[ultra thick,color=c,fill=c!10] (2,2) rectangle (1,1);
		\draw[ultra thick,color=c,fill=c!10] (1,2) -- (2,2) -- (2,3) -- (1,2);
		\draw[ultra thick,color=c,fill=c!10] (1,2) -- (2,3) -- (2,4) -- (1,3) -- (1,2);
		\oot{(0,0)}
	\end{scope}
\end{tikzpicture}

\pagebreak\section{Origami proof}

% <https://commons.wikimedia.org/wiki/File:Scissors_icon_black.svg>
\newcommand\scissorsicon{%
    \fill (9.6305,1.2561) .. controls (9.1986,0.7379) and (8.4222,0.7585) .. (7.8428,0.9834) .. controls (7.8428,0.9834) and (4.2034,2.3596) .. (4.2034,2.3596) .. controls (3.1422,1.7830) and (2.0956,1.9521) .. (2.0948,1.6698) .. controls (2.0942,1.4492) and (2.2969,1.4969) .. (2.2649,0.9226) .. controls (2.2342,0.3711) and (1.6571,-0.0353) .. (1.1130,0.0024) .. controls (0.5685,0.0008) and (0.0053,0.4197) .. (0.0022,0.9951) .. controls (-0.0367,1.5748) and (0.4523,2.1116) .. (1.0289,2.1479) .. controls (1.7046,2.2464) and (2.8994,1.8338) .. (3.4635,2.8376) .. controls (3.0477,3.6100) and (2.3061,3.6144) .. (1.6354,3.5777) .. controls (1.0786,3.5473) and (0.4104,3.7088) .. (0.1465,4.2568) .. controls (-0.1086,4.7820) and (0.1452,5.5277) .. (0.7348,5.6865) .. controls (1.3301,5.9029) and (2.1458,5.6248) .. (2.2852,4.9407) .. controls (2.3926,4.4134) and (2.0187,4.1669) .. (2.1614,3.9752) .. controls (2.2691,3.8305) and (2.9557,3.9121) .. (4.1769,3.3593) .. controls (4.1769,3.3593) and (8.1263,4.7384) .. (8.1263,4.7384) .. controls (8.6252,4.8761) and (9.2344,4.8696) .. (9.6745,4.3782) .. controls (9.6745,4.3782) and (5.5795,2.8440) .. (5.5795,2.8440) .. controls (5.5795,2.8440) and (9.6305,1.2561) .. (9.6305,1.2561) .. controls (9.6305,1.2561) and (9.6305,1.2561) .. (9.6305,1.2561)(1.5059,0.4369) .. controls (2.0812,0.7559) and (2.0368,1.6259) .. (1.4405,1.7961) .. controls (0.9013,1.9857) and (0.2101,1.4630) .. (0.3805,0.8776) .. controls (0.4798,0.3973) and (1.1121,0.2185) .. (1.5059,0.4369) .. controls (1.5059,0.4369) and (1.5059,0.4369) .. (1.5059,0.4369)(1.5413,3.9937) .. controls (2.1413,4.2132) and (2.0425,5.2275) .. (1.4335,5.3656) .. controls (1.0297,5.5071) and (0.4512,5.3069) .. (0.4090,4.8308) .. controls (0.3285,4.2521) and (1.0073,3.7270) .. (1.5413,3.9937) .. controls (1.5413,3.9937) and (1.5413,3.9937) .. (1.5413,3.9937);
}



% dashed = valley
% dashdotted = mountain
% thin solid = previous crease
% dotted = showing paper behind
% <https://en.wikipedia.org/wiki/Yoshizawa%E2%80%93Randlett_system>
1. \begin{tikzpicture}[baseline]
	\draw[thick,color=black,fill=e!15] (0,0) rectangle (2,2);
	\draw[thin,dashed] (0,1) -- (2,1);
	\draw[thin,->,looseness=0.8,out=225,in=135] (1,1.5) to (1,0.5);
\end{tikzpicture}
\quad
2. \begin{tikzpicture}[baseline]
	\draw[thick,color=black,fill=e!15] (0,1) -- (2,1) -- (2.25,0.25) -- (0.25,0.25) -- cycle;
	\draw[thick,color=black,fill=d!50] (0,0) rectangle (2,1);
	\draw[thin,dashed] (1,1) -- (1,0);
	\draw[thin,->,looseness=0.8,out=45,in=135] (0.5,0.5) to (1.5,0.5);
\end{tikzpicture}
\quad
3. \begin{tikzpicture}[baseline]
	\draw[thick,color=black,fill=d!50] (0,1) -- (0.9,1.1) -- (1,0.1) -- (0.1,0.2) -- cycle;
	\draw[thick,color=black,fill=e!15] (0,1) -- (1,1) -- (1.1,0.1) -- (0.1,0.1) -- cycle;
	\draw[thick,color=black,fill=d!50] (0,0) rectangle (1,1);
	\draw[thin,dashed] (1,1) -- (0,0);
	\draw[thin,->,looseness=0.8,out=0,in=90] (0.25,0.75) to (0.75,0.25);
\end{tikzpicture}
\quad
4. \begin{tikzpicture}[baseline]
	\draw[thick,color=black,fill=d!50] (0,0) -- (1.1,0.1) -- (1,1) -- cycle;
	\draw[thick,color=black,fill=d!50] (0,0) -- (1,0) -- (1.05,0.9) -- cycle;
	\draw[thick,color=black,fill=d!50] (0,0) -- (1,0) -- (1,1) -- cycle;
	\draw[thin,dashed] (0.3,0.3) -- (1,0.3);
	\draw[thin,->,looseness=0.8,out=225,in=135] (0.75,0.625) to (0.75,0.125);
\end{tikzpicture}
\quad
5. \begin{tikzpicture}[baseline]
	\draw[thick,color=black,fill=d!50] (0,0) -- (1,0) -- (1,0.3) -- (0.3,0.3) -- cycle;
	\draw[thick,color=black,fill=d!50] (0.3,0.3) -- (1,0.3) -- (1.05,-0.3) -- cycle; % 2
	\draw[thick,color=black,fill=d!50] (0.3,0.3) -- (1.1,0.3) -- (1,-0.4) -- cycle; % 1
	\draw[thin,->,looseness=0.8,out=45,in=315] (0.75,0.125) to (0.75,0.625);
\end{tikzpicture}
\quad
6. \begin{tikzpicture}[baseline]
	\draw[thick,color=black,fill=e!15] (0,0) rectangle (2,2);
	\draw[thin,shorten >=2,shorten <=2] (0,0.6) -- (2,0.6); % h 1
	\draw[thin,shorten >=2,shorten <=2] (0,1)   -- (2,1);   % h 2
	\draw[thin,shorten >=2,shorten <=2] (0,1.4) -- (2,1.4); % h 3
	\draw[thin,shorten >=2,shorten <=2] (0.6,0) -- (0.6,2); % v 1
	\draw[thin,shorten >=2,shorten <=2] (1,0)   -- (1,2);   % v 2
	\draw[thin,shorten >=2,shorten <=2] (1.4,0) -- (1.4,2); % v 3
	\draw[thin,shorten >=2,shorten <=2] (0,0)   -- (2,2);   % d 1
	\draw[thin,shorten >=2,shorten <=2] (0,2)   -- (2,0);   % d 2
	\draw[thin,dashdotted] (1.4,2) -- (0,1.4);
	\draw[thin,->,looseness=0.8,out=105,in=15] (1.1,1.7) to (0.3,2.1);
\end{tikzpicture}
\quad
7. \begin{tikzpicture}[baseline]
	\draw[thick,color=black,fill=e!15] (0,0) -- (0,1.4) -- (1.4,2) -- (2,2) -- (2,0) -- cycle; % y = 2/7x + 1.6
	\draw[thin,shorten >=2,shorten <=2] (0,0.6)     -- (2,0.6);    % h 1
	\draw[thin,shorten >=2,shorten <=2] (0,1)       -- (2,1);      % h 2
	\draw[thin,shorten >=2,shorten <=2] (0,1.4)     -- (2,1.4);    % h 3
	\draw[thin,shorten >=2,shorten <=2] (0.6,0)     -- (0.6,1.66); % v 1
	\draw[thin,shorten >=2,shorten <=2] (1,0)       -- (1,1.83);   % v 2
	\draw[thin,shorten >=2,shorten <=2] (1.4,0)     -- (1.4,2);    % v 3
	\draw[thin,shorten >=2,shorten <=2] (0,0)       -- (2,2);      % d 1
	\draw[thin,shorten >=2,shorten <=2] (0.42,1.58) -- (2,0);      % d 2
	\draw[thick,color=a] (1.4,2) -- (1.4,1.4) node[midway,right] {\tiny $\sidea$};
	\draw[thick,color=b] (1.4,1.4) -- (0,1.4) node[midway,below] {\tiny $\sideb$};
	\draw[thick,color=c] (0,1.4) -- (1.4,2) node[midway,above left] {\tiny $\sidec$};
	\begin{scope}[scale=0.5]
		\ninetydeg[90]{2.8}{2.8}
	\end{scope}
	\draw[thin,dashdotted] (0,1.4) -- (0.6,0);
	\draw[thin,dashdotted] (0.6,0) -- (2,0.6);
	\draw[thin,dashdotted] (2,0.6) -- (1.4,2);
	\draw[thin,->,looseness=0.8,out=195,in=105] (0.3,1.3) to (-0.1,0.1);
	\draw[thin,->,looseness=0.8,out=285,in=195] (0.7,0.3) to (1.9,-0.1);
	\draw[thin,->,looseness=0.8,out=15,in=285]  (1.7,0.7) to (2.1,1.9);
\end{tikzpicture}
\quad
8. \begin{tikzpicture}[baseline]
	% \draw[thin,->,looseness=0.8,out=15,in=105]  (0.3,2.1) to (1.1,1.7);
	% \draw[thin,->,looseness=0.8,out=105,in=195] (-0.1,0.1) to (0.3,1.3);
	% \draw[thin,->,looseness=0.8,out=195,in=285] (1.9,-0.1) to (0.7,0.3);
	% \draw[thin,->,looseness=0.8,out=285,in=15]  (2.1,1.9) to (1.7,0.7);
	\draw[thick,color=c,fill=c!10] (0,1.4) -- node[midway,above] {\small $\sidec$}
	                               (1.4,2) -- node[midway,right] {\small $\sidec$}
	                               (2,0.6) -- node[midway,below] {\small $\sidec$}
	                               (0.6,0) -- node[midway,left]  {\small $\sidec$}
	cycle; % y = 2/7x + 1.6
	\draw[thin] (0.34,0.6)  -- (2,0.6);     % h 1
	\draw[thin] (0.17,1)    -- (1.83,1);    % h 2
	\draw[thin] (0,1.4)     -- (1.66,1.4);  % h 3
	\draw[thin] (0.6,0)     -- (0.6,1.66);  % v 1
	\draw[thin] (1,0.17)    -- (1,1.83);    % v 2
	\draw[thin] (1.4,0.34)  -- (1.4,2);     % v 3
	\draw[thin] (0.42,0.42) -- (1.58,1.58); % d 1
	\draw[thin] (0.42,1.58) -- (1.58,0.42); % d 2
\end{tikzpicture}
\quad
9. \begin{tikzpicture}[baseline]
	\draw[thick,color=black,fill=e!15] (0.6,0) -- (0.6,1.4) -- (1.4,1.4) -- (1.4,2) -- (2,2) -- (2,0) -- cycle;
	\begin{scope}[shift={(-0.1,0.1)}]
		\draw[thick,color=black,fill=e!15] (0,1.4) -- (0,2) -- (1.4,2) -- (1.4,1.4) -- cycle;
		\draw[thin,shorten >=2,shorten <=2] (0.6,1.4) -- (0.6,2); % v 1
		\draw[thin,shorten >=2,shorten <=2] (1,1.4)   -- (1,2);   % v 2
	\draw[thin,shorten >=2,shorten <=2] (0,2)   -- (0.6,1.4);   % d 2
		\draw[thin,shorten >=2,shorten <=2] (1.4,2) -- (0,1.4);
	\end{scope}
	\begin{scope}[shift={(-0.1,0)}]
		\draw[thick,color=black,fill=e!15] (0,0) -- (0,1.4) -- (0.6,1.4) -- (0.6,0) -- cycle;
		\draw[thin,shorten >=2,shorten <=2] (0,0.6) -- (0.6,0.6); % h 1
		\draw[thin,shorten >=2,shorten <=2] (0,1)   -- (0.6,1);   % h 2
	\draw[thin,shorten >=2,shorten <=2] (0,0)   -- (0.6,0.6);   % d 1
	\end{scope}
	\draw[thin,shorten >=2,shorten <=2] (0.6,0.6) -- (2,0.6);   % h 1
	\draw[thin,shorten >=2,shorten <=2] (0.6,1)   -- (2,1);     % h 2
	\draw[thin,shorten >=2,shorten <=2] (0.6,1.4) -- (2,1.4);   % h 3
	\draw[thin,shorten >=2,shorten <=2] (0.6,0)   -- (0.6,1.4); % v 1
	\draw[thin,shorten >=2,shorten <=2] (1,0)     -- (1,1.4);   % v 2
	\draw[thin,shorten >=2,shorten <=2] (1.4,0)   -- (1.4,1.4); % v 3
	\draw[thin,shorten >=2,shorten <=2] (0.6,0.6) -- (2,2);     % d 1
	\draw[thin,shorten >=2,shorten <=2] (0.6,1.4) -- (2,0);     % d 2
	\draw[thin,shorten >=2,shorten <=2] (0,1.4)   -- (0.4,0);
	\draw[thin,shorten >=2,shorten <=2] (0.6,0)   -- (2,0.6);
	\draw[thin,shorten >=2,shorten <=2] (2,0.6)   -- (1.4,2);
	\begin{scope}[shift={(1.2,1.6)},rotate=180,scale=0.05] \scissorsicon \end{scope}
	\begin{scope}[shift={(0.4,1)},rotate=270,scale=0.05] \scissorsicon \end{scope}
\end{tikzpicture}
\quad
10. \begin{tikzpicture}[baseline]
	% \draw[thick,color=black,fill=e!15] (0.6,0) -- (0.6,1.4) -- (1.4,1.4) -- (1.4,2) -- (2,2) -- (2,0) -- cycle;
	\draw[thick,color=a,fill=a!10] (1.4,1.4) -- node[midway,left] {\small $\sidea$}
	                               (1.4,2) -- node[midway,above] {\small $\sidea$}
	                               (2,2) -- node[midway,right] {\small $\sidea$}
	                               (2,1.4) -- cycle;
	\draw[thick,color=b,fill=b!10] (0.6,0) -- node[midway,left] {\small $\sideb$}
	                               (0.6,1.4) -- (2,1.4) -- node[midway,right] {\small $\sideb$}
	                               (2,0) -- node[midway,below] {\small $\sideb$}
	                               cycle;
	\draw[thin,shorten >=2,shorten <=2] (0.6,0.6) -- (2,0.6); % h 1
	\draw[thin,shorten >=2,shorten <=2] (0.6,1)   -- (2,1);   % h 2
	% \draw[thin,shorten >=2,shorten <=2] (0.6,1.4) -- (2,1.4); % h 3
	% \draw[thin,shorten >=2,shorten <=2] (0.6,0) -- (0.6,1.4); % v 1
	\draw[thin,shorten >=2,shorten <=2] (1,0)   -- (1,1.4);   % v 2
	\draw[thin,shorten >=2,shorten <=2] (1.4,0) -- (1.4,1.4); % v 3
	\draw[thin,shorten >=2,shorten <=2] (0.6,0.6)   -- (2,2);   % d 1
	\draw[thin,shorten >=2,shorten <=2] (0.6,1.4)   -- (2,0);   % d 2
	\draw[thin,shorten >=2,shorten <=2] (0.6,0) -- (2,0.6);
	\draw[thin,shorten >=2,shorten <=2] (2,0.6) -- (1.4,2);
\end{tikzpicture}

\par\bigskip
Fold origami paper in half twice (1--2), then along the diagonal (3). Fold the top corner down---the height of this crease determines the ratio of $a$ to $b$ (4). Unfold all (5). Fold the top corner back along the diagonal---this is the first triangle (6). Repeat with the other three corners (7). The remaining square has area $c^2$ (8). Unfold, and cut out four triangles from the top and left (9). The remaining paper has area $a^2+b^2$ (10).

\begin{itemize}
	\item Posted to YouTube by Vi Hart in 2011.
\end{itemize}

\end{document}