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\begin{tikzpicture}
    \tikzset{mypoints/.style={fill=white,draw=black}}
    \tikzstyle{help lines}+=[very thin]


    \coordinate[label={[label distance=0.5em]180:$A$}]
        (A) at (120:5);
    \coordinate[label={[label distance=0.5em]180:$B$}]
        (B) at (210:5);
    \coordinate[label={[label distance=0.5em]  0:$C$}]
        (C) at (335:5);
    \coordinate[label={[label distance=0.5em]  0:$D$}]
        (D) at ( 40:5);
    \coordinate[label={[label distance=0.5em]-90:$M$}]
        (M) at (intersection of A--C and B--D);

    \draw (A) -- (B) -- (C) -- (D) -- (A);
    \draw (A) -- (C) (B) -- (D);
\end{tikzpicture}

In the example above, how to construct a point E on AC so that ∠ADE = ∠BDC?

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1 Answer

up vote 6 down vote accepted

This looks like a job for the excellent tkz-euclide package. The syntax takes a bit of getting used to, but it's very powerful:

\documentclass{standalone}
\usepackage{tkz-euclide}
\usetkzobj{all}

\begin{document}
\begin{tikzpicture}

% Define the known points
\tkzDefPoint(120:5){A} 
\tkzDefPoint(210:5){B}
\tkzDefPoint(335:5){C}
\tkzDefPoint(40:5){D}

% Find the intersection between AC and BD, save point as M
\tkzInterLL(A,C)(B,D)
\tkzGetPoint{M}

% Measure the angle BDC, save as \angleBDC
\tkzFindAngle(B,D,C)
\tkzGetAngle{angleBDC}

% Rotate A by \angleBDC around D, save point as E'
\tkzDefPointBy[rotation=center D angle \angleBDC](A)
\tkzGetPoint{E'}

%Find the intersection between AC and DE', save point as E
\tkzInterLL(A,C)(D,E')
\tkzGetPoint{E}

\tkzDrawSegment(A,C)
\tkzDrawSegment(B,D)
\tkzDrawSegment(D,E)

\tkzDrawPolygon(A,B,C,D)


\tkzDrawPoints(A,B,C,D,M,E)
\tkzLabelPoints[above](A,D)
\tkzLabelPoints[below](B,C,M)
\tkzLabelPoints[below left](E)
\end{tikzpicture}
\end{document}
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Is there a way to use \tkzLabelAngle[pos = 0.8](B,A,C){angleBAC} for writing that value near the angle? –  Arne Aug 24 '13 at 17:52
    
\tkzGetAngle{angleBAC}; \tkzLabelAngle[pos = 0.8](B,A,C){\tkzAngleResult}; –  Arne Aug 24 '13 at 18:20
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