38 lines
1.2 KiB
TeX
38 lines
1.2 KiB
TeX
|
|
||
|
|
\documentclass{beamer}
|
||
|
|
\usetheme{Berlin}
|
||
|
|
\usepackage[utf8]{inputenc}
|
||
|
|
\usepackage{polski}
|
||
|
|
\author{Grzegorz Adamski}
|
||
|
|
\usepackage{tikz}
|
||
|
|
\useinnertheme[shadow=true]{rounded}
|
||
|
|
\useoutertheme{infolines}
|
||
|
|
\usecolortheme{wolverine}
|
||
|
|
\setbeamercolor{alerted text}{fg=red}
|
||
|
|
\title[Pitagoras]{Dowód twierdzenia Pitagorasa}
|
||
|
|
\date{09.11.2017}
|
||
|
|
\begin{document}
|
||
|
|
\maketitle
|
||
|
|
\begin{frame}{This is dowód}
|
||
|
|
\begin{center}
|
||
|
|
\begin {tikzpicture}
|
||
|
|
\draw[blue] (0 ,0)--(1 ,2)--(5 ,0)--(0,0);
|
||
|
|
\draw[blue] (1 ,2)--(1,0);
|
||
|
|
\fill ( 5 , 0 ) circle[radius=2pt];
|
||
|
|
\node [below right] at ( 5 , 0 ) {$B$};
|
||
|
|
\fill ( 0 , 0 ) circle[radius=2pt] ;
|
||
|
|
\node [below right] at ( 0 , 0 ) {$A$};
|
||
|
|
\fill ( 1 , 2 ) circle[radius=2pt] ;
|
||
|
|
\node [above right] at ( 1 , 2 ) {$C$};
|
||
|
|
\fill ( 1 , 0 ) circle[radius=2pt] ;
|
||
|
|
\node [below right] at ( 1 , 0 ) {$D$};
|
||
|
|
\end{tikzpicture}
|
||
|
|
\end{center}
|
||
|
|
|
||
|
|
Trójkąty $ADC$, $BCD$ i $ABC$ są podobne, zatem $|AD|=a$, $|DC|=ab$, $|DB|=ab^2$, $|AC|=c$, $|BC|=cb$. Pole trójkąta $ABC$ jest równe sumie pól trójkątów $ADC$ i $BCD$, zatem:
|
||
|
|
\[\frac{a\cdot ab}{2}+\frac{ab\cdot ab^2}{2}=\frac{c\cdot cb}{2}.\]
|
||
|
|
Po skróceniu otrzymujemy:
|
||
|
|
\[a^2+b^2=c^2.\]
|
||
|
|
\end{frame}
|
||
|
|
|
||
|
|
\end{document}
|