37 lines
1.2 KiB
TeX
37 lines
1.2 KiB
TeX
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\documentclass{beamer}
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\usetheme{Berlin}
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\usepackage[utf8]{inputenc}
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\usepackage{polski}
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\author{Grzegorz Adamski}
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\usepackage{tikz}
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\useinnertheme[shadow=true]{rounded}
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\useoutertheme{infolines}
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\usecolortheme{wolverine}
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\setbeamercolor{alerted text}{fg=red}
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\title[Pitagoras]{Dowód twierdzenia Pitagorasa}
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\date{09.11.2017}
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\begin{document}
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\maketitle
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\begin{frame}{This is dowód}
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\begin{center}
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\begin {tikzpicture}
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\draw[blue] (0 ,0)--(1 ,2)--(5 ,0)--(0,0);
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\draw[blue] (1 ,2)--(1,0);
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\fill ( 5 , 0 ) circle[radius=2pt];
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\node [below right] at ( 5 , 0 ) {$B$};
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\fill ( 0 , 0 ) circle[radius=2pt] ;
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\node [below right] at ( 0 , 0 ) {$A$};
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\fill ( 1 , 2 ) circle[radius=2pt] ;
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\node [above right] at ( 1 , 2 ) {$C$};
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\fill ( 1 , 0 ) circle[radius=2pt] ;
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\node [below right] at ( 1 , 0 ) {$D$};
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\end{tikzpicture}
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\end{center}
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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:
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\[\frac{a\cdot ab}{2}+\frac{ab\cdot ab^2}{2}=\frac{c\cdot cb}{2}.\]
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Po skróceniu otrzymujemy $a^2+(ab)^2=c^2$, czyli twierdzenie Pitagorasa dla trójkąta $ADC$.
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\end{frame}
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\end{document} |