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Aleksandra Jonas 2019-01-29 14:01:37 +00:00
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\documentclass[]{beamer}
\usepackage{tikz}
\usepackage[polish]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usetheme{EastLansing}
\begin{document}
\author{Aleksandra Jonas}
\title{Twierdzenie Telesa}
\institute{Adam Mickiewicz University}
\date{21 stycznia 2019}
\begin{frame}[plain]
\maketitle
\end{frame}
\begin{frame}
\begin{center}
\frametitle{Tales}
\begin{figure}
\includegraphics[scale=0.75]{j:/Downloads/tales}
\end{figure}
\end{center}
\end{frame}
\begin{frame}[plain]
\begin{center}
\begin{tikzpicture}[]
\draw
[black, dotted] (-2,0)--(4,0)--(3,3)--(2,6);
\draw
[pink] (1,-1)--(5,3)
[pink] (-1,-1)--(5,5)
[red, very thick]
(4,0) node [below, right]{A}
(2,0) node [below]{B}
(0,0) node [below] {C}
(3.5,1.5) node [right]{D}
(3,3) node [right]{E}
[black, very thick]
(5,2) node [above] {k}
(5,4) node [above, left] {m}
(-1,4) node [above] {k II m}
[pink]
(4,0) node {•}
(2,0) node {•}
(3,3) node {•}
(3.5,1.5) node {•}
(0,0) node {•};
\end{tikzpicture}
\end{center}
\end{frame}
\begin{frame}
\begin{itemize}
\item Jeżeli ramiona kąta przecięte są prostymi równoległymi, to: \pause
\item $\dfrac{|AD|}{|AB|}=$\pause $\dfrac{|BC|}{|DE|}$
\end{itemize}
\end{frame}
\begin{frame}
\begin{center}
Dziękuję za uwagę!
\end{center}
\end{frame}
\end{document}