# PWWs-01-07

(Last updated: 2016-10-31 10:12) A Pythagorean Theorem: $$a\cdot a' = b\cdot b'+c\cdot c'$$

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 \documentclass[tikz, border=2pt]{standalone} \usepackage{amssymb} \usetikzlibrary{calc} \usetikzlibrary{math} \usetikzlibrary{arrows.meta} \begin{document} \begin{tikzpicture}[line width=1pt] \tikzmath{ \b = 7; \c = 12; \a = sqrt(\b*\b + \c*\c); \l = 7; \b = \b / \a * \l; \c = \c / \a * \l; \a = \l; \B = asin(\b / \a); \C = asin(\c / \a); } \def\template#1; { \coordinate (O) at #1; \coordinate (A) at ($(\C:\b)+(O)$); \coordinate (B) at ($(\a, 0)+(O)$); \coordinate (B1) at ($(0.3*\a, 0)+(O)$); \coordinate (A1) at ($(\C:0.7*\b)+(B1)$); \draw (O) -- node[below, near start] {$a$} (B) -- node[above right, near end] {$c$} (A) -- node[left] {$b$} cycle; \draw (B1) -- node[left] {$b'$} (A1); \node[above] () at ($(B1)!0.45!(B)$) {$a'$}; \node[below] () at ($(A1)!0.3!(B)$) {$c'$}; \draw[thin] (A) ++(-\B:0.04*\l) -- ++({-90-\B}:0.04*\l) -- ++({180-\B}:0.04*\l); \draw[thin] (A1) ++(-\B:0.04*\l) -- ++({-90-\B}:0.04*\l) -- ++({180-\B}:0.04*\l); } \template{(0, 4)}; \template{(0, 0)}; \coordinate (P) at ($(A1)+(0, -0.7*\b*\c/\a)$); \draw[thin] (A1) -- (P); \draw[thin] (P) ++(0.04*\l, 0) -- ++(0, 0.04*\l) -- ++(-0.04*\l, 0); \draw[thin] (P) -- +(0, -0.05*\l) node[coordinate, near end] (p1) {}; \draw[thin] (B1) -- +(0, -0.05*\l) node[coordinate, near end] (p2) {}; \draw[thin] (B) -- +(0, -0.05*\l) node[coordinate, near end] (p3) {}; \node (x) at ($(p1)!0.5!(p2)$) {$x$}; \node (y) at ($(p1)!0.5!(p3)$) {$y$}; \draw[-{Latex[length=2mm]}, thin] (x) -- (p1); \draw[-{Latex[length=2mm]}, thin] (x) -- (p2); \draw[-{Latex[length=2mm]}, thin] (y) -- (p1); \draw[-{Latex[length=2mm]}, thin] (y) -- (p3); \node[below=5mm, align=center] () at ($(O)!0.5!(B)$) { $\displaystyle\frac{x}{b}=\displaystyle\frac{b'}{a} \Rightarrow a\cdot x = b\cdot b';$\\ $\displaystyle\frac{y}{c}=\displaystyle\frac{c'}{a} \Rightarrow a\cdot y = c\cdot c';$\\ $\therefore\ a\cdot a'=a\cdot(x+y)=b\cdot b'+c\cdot c'.$ }; \end{tikzpicture} \end{document}