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+
+\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}
\ No newline at end of file