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%opening
\makeatletter
\newcommand{\stevilkadn}{6}
\newcommand{\snovdn}{Skladnost trikotnikov }
\newcommand{\cm}[1]{\SI{#1}{\centi\meter}}
\newcommand{\xslalph}[1]{\expandafter\@xslalph\csname c@#1\endcsname}
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\renewcommand\abstractname{Povzetek}
\date{17. september 2020}
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\newcommand{\iic}{I\textsuperscript{2}C }
% \patchcmd{\thebibliography}{\section*{\refname}}{}{}{}
\title{%
\snovdn --- \stevilkadn. domača naloga \\
\large Matematika, Gimnazija Bežigrad}
\author{\begin{tabular}{rl}
\textbf{Profesor:} & prof. Vilko Domajnko \\
\textbf{Avtor:} & Anton Luka Šijanec, 2. a
% \textbf{Avtor:} & Anton Luka Šijanec \\ & Member 2 \\ & Member 3
\end{tabular}}
% \everymath{\displaystyle} % https://tex.stackexchange.com/a/32847/212260
\begin{document}
\maketitle
\begin{abstract}
Ta dokument obsega naloge, naročene dijakom 16. septembra 2020, in njihove
rešitve, ki sem jih spisal sam. Naloge obsegajo snov \textit{\snovdn}in so
iz učbenika in vaj/zbirke nalog.
\end{abstract}
% \tableofcontents
\section{Učbenik \textit{Matematika 1}: Stran 19}
\begin{enumerate}[label=\textbf{\arabic*.}]
\item Načrtaj kote, velike \ang[parse-numbers=false]{\textbf{105}}, \ang{150}, \ang{15}, \ang{135} in \ang{165}.
\begin{center}
\begin{tikzpicture}
\tkzDefPoint(-5,0){A} %%% kot 90 stopinj ˇˇˇ
\tkzDefPoint(5,0){B}
\tkzDefPoint(0,0){CHEATING0} \tkzDefPoint(0.05,0){CHEATING1} \tkzDefPoint(-0.05,0){CHEATING2}
\tkzInterCC(A,B)(B,A) \tkzGetPoints{C}{D}
\tkzDrawPoints(A) \tkzCompass(A,C) \tkzCompass(B,C) \tkzCompass(A,CHEATING1) \tkzCompass(B,CHEATING2) \tkzCompass(A,B)
\tkzDrawLine(A,B) \tkzDrawLine(CHEATING0,C) \tkzMarkAngle[mark=none](B,CHEATING0,C)
\tkzMarkRightAngles(B,CHEATING0,C)
%%% kot 60 stopinj ˇˇˇ
\tkzInterLC(CHEATING0,C)(CHEATING0,A) \tkzGetPoints{E}{F} \tkzCompass(CHEATING0,F)
\tkzInterCC(CHEATING0,F)(F,CHEATING0) \tkzGetPoints{G}{H} \tkzCompass(CHEATING0,G) \tkzCompass(F,G)
\tkzDrawLine(CHEATING0,G) \tkzMarkAngle[color=magenta, mark=none](F,CHEATING0,G)
%%% kot 60/2 stopinj ˇˇˇ
\tkzDrawLine(F,G) \tkzDefMidPoint(F,G) \tkzGetPoint{I} \tkzCompass(F,I) \tkzCompass(G,I)
\tkzMarkAngle[color=cyan,mark=none, size=1.1cm](F,CHEATING0,I)
%%% kot 60/2/2 stopinj = 15 stopinj ˇˇˇ
\tkzDefMidPoint(F,I) \tkzGetPoint{J} \tkzCompass(F,J) \tkzCompass(I,J)
\tkzMarkAngle[color=yellow,mark=none, size=1.2cm](F,CHEATING0,J)
%%% kot 105 stopinj = kot 90 stopinj + kot 15 stopinj ˇˇˇ
\tkzMarkAngle[color=red,mark=none, size=0.9cm](B,CHEATING0,J) \tkzDrawLine(CHEATING0,J)
\end{tikzpicture}
\end{center}
\item Načrtaj trikotnik, ki ima stranice dolge \SI{6}{\centi\meter}, \SI{4}{\centi\meter} in \SI{4,5}{\centi\meter}.
\item Načrtaj trikotnik z danimi podatki.
\begin{multicols}{4}
\noindent
\begin{enumerate}[label=\textbf{\xslalph*)}]
\item $\mathbf{c=}\SI[parse-numbers=false, parse-units=false]{\mathbf{5}}{\mathbf{\centi\meter}}$
$\mathbf{b=}\SI[parse-numbers=false, parse-units=false]{\mathbf{4}}{\mathbf{\centi\meter}}$
$\mathbf{\alpha=}\ang[parse-numbers=false, parse-units=false]{\mathbf{75}}$
\item $a=\SI{6}{\centi\meter}$
$\beta=\ang{60}$
$\gamma=\ang{45}$
\item $c=\SI{5,5}{\centi\meter}$
$a=\SI{4}{\centi\meter}$
$\gamma=\ang{75}$
\item $b=\SI{5}{\cm}$
$c=\SI{6,5}{\centi\meter}$
$\beta=\ang{60}$
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\noindent
\begin{center}
\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(5,0){B}
\tkzDefPoint(4,0){b}
\tkzDrawPoints(A) \tkzCompass(A,b) \tkzDrawLine(A,B) \tkzCompass(A,b)
\tkzDrawSegment[dim={$\SI{4,0}{\centi\meter}$,6pt,transform shape}](A,b)
\tkzInterCC(A,b)(b,A) \tkzGetPoints{D}{d} \tkzCompass(A,D) \tkzCompass(B,D) \tkzInterCC(D,A)(A,D) \tkzGetPoints{e}{E}
\tkzCompass(A,E) \tkzCompass(D,E) \tkzDefMidPoint(E,D) \tkzGetPoint{F} \tkzCompass(D,F) \tkzCompass(E,F)
\tkzDefMidPoint(F,D) \tkzGetPoint{C} \tkzCompass(F,C) \tkzCompass(D,C) \tkzDrawLine(A,C) \tkzDrawLine(B,C)
\tkzLabelLine[below left](B,C){$a$} \tkzLabelLine[below right](C,A){$b$} \tkzLabelLine[below](A,B){$c$}
\tkzLabelPoints(A) \tkzLabelPoints[below left](B) \tkzLabelPoints[above right](C)
\tkzMarkAngle[mark=none](B,A,C) \tkzLabelAngle[pos=1.2](B,A,C){$\alpha$}
\end{tikzpicture}
\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(6,0){B}
\tkzDefPoint(4,0){a}
\tkzDefPoint(4.5+6,0){b}
\tkzDrawPoint(A)
\tkzCompass(A,B)
\tkzInterCC(A,a)(B,b) \tkzGetPoints{C}{c} \tkzCompass(A,C) \tkzCompass(B,C)
\tkzDrawLine(A,B) \tkzDrawLine(B,C) \tkzDrawLine(C,A)
\end{tikzpicture}
\end{center}
\end{multicols}
\item Načrtaj štirikotnik z danimi podatki. Pomagaj si s trikotnika, na katera diagonala razdeli štirikotnik
\begin{multicols}{6}
\noindent
\begin{enumerate}[label=\textbf{\xslalph*)}]
\item $\mathbf{a=}\SI[parse-numbers=false, parse-units=false]{\mathbf{5}}{\mathbf{\centi\meter}}$
$\mathbf{b=}\SI[parse-numbers=false, parse-units=false]{\mathbf{3}}{\mathbf{\centi\meter}}$
$\mathbf{c=}\SI[parse-numbers=false, parse-units=false]{\mathbf{4}}{\mathbf{\centi\meter}}$
$\mathbf{d=}\SI[parse-numbers=false, parse-units=false]{\mathbf{4}}{\mathbf{\centi\meter}}$
$\mathbf{f=}\SI[parse-numbers=false, parse-units=false]{\mathbf{5}}{\mathbf{\centi\meter}}$
\item $a=\SI{6}{\centi\meter}$
$b=\SI{5}{\centi\meter}$
$d=\SI{4}{\centi\meter}$
$\alpha=\ang{120}$
$\delta=\ang{90}$
\item $b=\SI{6}{\centi\meter}$
$c=\cm{2}$
$d=\cm{5}$
$\gamma=\ang{105}$
$\delta=\ang{75}$
\item $a=\cm{7}$
$b=\cm{4}$
$c=\cm{5}$
$e=\cm{6}$
$f=\cm{5,5}$
\item $d=c=\cm{3}$
$a=d=\cm{5}$
$f=\cm{5,5}$
\item $a=\cm{5}$
$d=\cm{4}$
$e=\cm{6}$
$f=\cm{7}$
$\beta=\ang{60}$
\end{enumerate}
\end{multicols}
% \begin{center}
Štirikotnikov še nismo definirali. (-; Problem je v tem, da ne vem, med katerimi pari točk so določene stranice.
% \end{center}
\end{enumerate}
\section{Zbirka nalog \textit{Matematika 1}: Stran 16}
\begin{enumerate}[label=\textbf{\arabic*.}]
\setcounter{enumi}{71}
\item Trikotnika $ABC$ in $A'B'C$ sta skladna. Točki $D$ in $D'$ ležita na stranicah $AB$ oziroma $A'B'$ tako, da sta kota $DCB$ in $D'C'B'$ skladna. Dokaži, da je $|AD|=|A'D'|$
\item Na sliki je $|AE|=|BG|$, kota $BAC$ in $FEG$ sta skladna, prav tako sta skladna kota $CBA$ in $EGF$. Dokaži, da sta trikotnika $ABC$ in $EGF$ skladna.
\item Točke $A, B, C, E$ in $F$ ležijo, kot kaže slika: $|AB|=|BC|, AF \bot BE, CE \bot BF$. Dokaži, da je $|AF|=|CE|$.
\begin{multicols}{2}
\begin{center}
\begin{tikzpicture}
\tkzDefPoint(0,0){B}
\tkzDefPoint(-5,0){C}
\tkzDefPoint(210:5){A}
%\tkzDefPoint(210:6){E}
\tkzDefLine[orthogonal=through A](B,A) \tkzGetPoint{f}
\tkzInterLL(A,f)(B,C)
\tkzGetPoint{F}
\tkzDefMidPoint(A,F) \tkzGetPoint{M}
\tkzDefLine[orthogonal=through M](B,F) \tkzGetPoint{e}
\tkzInterLL(C,e)(B,A) \tkzGetPoint{E}
\tkzDrawPoints(A,B,C,F,E)
\tkzLabelPoints(B,E,A) \tkzLabelPoints[above right](F,C)
\tkzDrawLine(B,F) \tkzDrawLine[add = 0 and 0](C,E)
\tkzDrawLine(B,E) \tkzDrawLine[add = 0 and 0](A,F)
\tkzMarkSegments[mark=|](C,B A,B) \tkzMarkSegments[mark=||](F,A E,C)
\tkzMarkAngles[mark=|||](C,B,A) \tkzLabelAngle[pos=2](C,B,A){$\beta$}
\tkzMarkRightAngles(B,A,F E,C,B)
\end{tikzpicture}
\end{center}
Trikotnika $ABF$ in $EBC$ sta skladna zaradi dveh skladnih kotov in ene stranice, zato so vse njune stranice enako dolge.
\end{multicols}
\end{enumerate}
\section{Zaključek}
Ta dokument je informativne narave in se lahko še spreminja. Najnovejša različica, torej PDFji in
\hologo{LaTeX}
izvorna koda, zgodovina sprememb in prejšnje različice, je na voljo mojem šolskem Git repozitoriju na
\url{https://github.com/sijanec/sola-gimb-2} v mapi
\href{https://github.com/sijanec/sola-gimb-2/tree/master/mat/domace_naloge/\stevilkadn/}{/mat/domace\_naloge/\stevilkadn/}. Povezava za ogled zadnje različice tega dokumenta v PDF obliki je \url{http://razor.arnes.si/~asija3/files/sola/gimb/2/mat/domace_naloge/\stevilkadn/dokument.pdf} in/ali \url{https://github.com/sijanec/sola-gimb-2/raw/master/mat/domace_naloge/\stevilkadn/dokument.pdf}.
\if\razhroscevanje1
\section{Razhroščevalne informacije}
Te informacije so generirane, ker je omogočeno razhroščevanje. Prej objavo dokumenta izklopite razhroščevanje. To naredite tako, da nastavite ukaz \texttt{razhroscevanje} na 0 v začetku dokumenta.
Grafi imajo natančnost \functionSamples\space točk na graf.
Konec generiranja dokumenta: \input|"date -Ins"
\fi
% \item $$$$
\end{document}
