PWWs-01-07

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

PWWs-01-07
\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}