Beweis. $ \begin{tikzpicture}[%scale=0.7, font=\footnotesize, background rectangle/.style={draw=none, fill=black!1, rounded corners}, %show background rectangle, Punkt/.style 2 args={ label={[#1]:$#2$} }, Dreieck/.style={thick}, ] \def\bew{1} % Gegebene Größen \pgfmathsetmacro{\a}{3.5} \pgfmathsetmacro{\b}{7} \pgfmathsetmacro{\c}{7.5} \pgfmathsetmacro{\s}{0.5*(\a+\b+\c)} \pgfmathsetmacro{\F}{sqrt( \s*(\s-\a)*(\s-\b)*(\s-\c) )} \pgfmathsetmacro{\Alpha}{acos((\b^2+\c^2-\a^2)/(2*\b*\c))} % \pgfmathsetmacro{\Beta}{acos((\a^2+\c^2-\b^2)/(2*\a*\c))} % \pgfmathsetmacro{\Gamma}{acos((\a^2+\b^2-\c^2)/(2*\a*\b))} % % Inkreisradius und -position \pgfmathsetmacro{\r}{\F/\s} \pgfmathsetmacro{\AI}{\c*sin(0.5*\Beta)/cos(0.5*\Gamma)} \pgfmathsetmacro{\AI}{\b*sin(0.5*\Gamma)/cos(0.5*\Beta)} \pgfmathsetmacro{\BI}{\a*sin(0.5*\Gamma)/cos(0.5*\Alpha)} \pgfmathsetmacro{\BI}{\c*sin(0.5*\Alpha)/cos(0.5*\Gamma)} \pgfmathsetmacro{\CI}{\a*sin(0.5*\Beta)/cos(0.5*\Alpha)} \pgfmathsetmacro{\CI}{\b*sin(0.5*\Alpha)/cos(0.5*\Beta)} % Winkelhalbierende \pgfmathsetmacro{\wa}{2*\F/(\a*cos(0.5*(\Beta-\Gamma)))} \pgfmathsetmacro{\wb}{2*\F/(\b*cos(0.5*(\Gamma-\Alpha)))} \pgfmathsetmacro{\wc}{2*\F/(\c*cos(0.5*(\Alpha-\Beta)))} % Dreieckskonstruktion \coordinate[label=below:$A$] (A) at (0,0); \coordinate[label=below:$B$] (B) at (\c,0); \coordinate[label=$C$] (C) at (\Alpha:\b); \draw[Dreieck, local bounding box=dreieck] (A) -- (B) -- (C) --cycle; % Winkelhalbierende \ifnum\bew=1 \pgfmathsetlengthmacro\w{4mm} \else \pgfmathsetlengthmacro\w{7mm} \fi \coordinate[label={[above=\w]:$I$}] (I) at (0.5*\Alpha:\AI); \draw[] (A) -- +(0.5*\Alpha:\wa) coordinate[label=45:$W_a$](Wa); \draw[] (B) -- +(180-0.5*\Beta:\wb) coordinate[label=135:$W_b$](Wb); \draw[] (C) -- +(-\Beta-0.5*\Gamma:\wc) coordinate[label=-90:$W_c$](Wc); % Inkreis \draw[] (I) circle[radius=\r]; % Außenwinkel \draw[] (A) -- +(-8mm,0) coordinate[](As); \draw[] (B) -- +(8mm,0) coordinate[](Bs); \draw[] (C) -- +(\Alpha:8mm) coordinate[](Cs); \draw pic [draw, angle radius=6mm, % pic text={$\Winkel$}, pic text options={}, "$\alpha'$", double, ] {angle =C--A--As}; \draw pic [draw, angle radius=6mm, % pic text={$\Winkel$}, pic text options={}, "$\beta'$", double, ] {angle =Bs--B--C}; \draw pic [draw, angle radius=6mm, % pic text={$\Winkel$}, pic text options={}, "$\gamma'$", double, ] {angle =B--C--Cs}; % Winkel \draw pic [draw, angle radius=11mm, angle eccentricity=1.3, % pic text={$\Winkel$}, pic text options={}, "$\tfrac\alpha2$", ] {angle =I--A--C}; \draw pic [draw, angle radius=13mm, angle eccentricity=1.3, % pic text={$\Winkel$}, pic text options={}, "$\tfrac\alpha2$", ] {angle =B--A--I}; \draw pic [draw, angle radius=4mm, angle eccentricity=1.6, % pic text={$\Winkel$}, pic text options={}, "$\tfrac\beta2$", ] {angle =C--B--I}; \draw pic [draw, angle radius=5mm, angle eccentricity=1.5, % pic text={$\Winkel$}, pic text options={}, "$\tfrac\beta2$", ] {angle =I--B--A}; \draw pic [draw, angle radius=3mm, angle eccentricity=1.6, % pic text={$\Winkel$}, pic text options={}, "$\tfrac\gamma2$", ] {angle =A--C--I}; \draw pic [draw, angle radius=3.5mm, angle eccentricity=1.55, % pic text={$\Winkel$}, pic text options={}, "$\tfrac\gamma2$", ] {angle =I--C--B}; %\def\bew{0} \ifnum\bew=1 \tikzset{ bew/.style={} } \def\x{x} \def\y{y} \def\z{z} \else \def\x{\tfrac{\gamma'}{2}} \def\y{\tfrac{\beta'}{2}} \def\z{\tfrac{\alpha'}{2}} \tikzset{ bew/.style={red, angle radius=7.5mm, angle eccentricity=0.7, } } \fi % Winkel an I \draw pic [draw, % pic text={$\Winkel$}, pic text options={}, "$\x$", bew ] {angle =Wb--I--A}; \draw pic [draw, % pic text={$\Winkel$}, pic text options={}, "$\y$", bew ] {angle =A--I--Wc}; \draw pic [draw, % pic text={$\Winkel$}, pic text options={}, "$\z$", bew ] {angle =Wc--I--B}; \draw pic [draw, % pic text={$\Winkel$}, pic text options={}, "$\x$", bew ] {angle =B--I--Wa}; \draw pic [draw, % pic text={$\Winkel$}, pic text options={}, "$\y$", bew ] {angle =Wa--I--C}; \draw pic [draw, % pic text={$\Winkel$}, pic text options={}, "$\z$", bew ] {angle =C--I--Wb}; %% Punkte \foreach \P in {Wa, Wb, Wc, I} \draw[fill=black!1] (\P) circle (1.5pt); \end{tikzpicture} $ Dem Schaubild entliest man das lineare Gleichungssystem $\def\gl#1{\textsf{(#1)}} \begin{array}{l l} x+y +\dfrac\beta2 +\dfrac\gamma2 =180^\circ \\[1em] \Leftrightarrow~ x+y = 180^\circ -\dfrac{\beta+\gamma}{2} = 180^\circ -\dfrac{180^\circ -\alpha}{2} & \Leftrightarrow~ \gl{1}~ x+y = 90^\circ +\dfrac\alpha2 \\[1em] % x+z +\dfrac\alpha2 +\dfrac\gamma2 =180^\circ \\[1em] \Leftrightarrow~ x+z = 180^\circ -\dfrac{\alpha+\gamma}{2} = 180^\circ -\dfrac{180^\circ -\beta}{2} & \Leftrightarrow~ \gl{2}~ x+z = 90^\circ +\dfrac\beta2 \\[1em] % y+z +\dfrac\alpha2 +\dfrac\beta2 =180^\circ \\[1em] \Leftrightarrow~ y+z = 180^\circ -\dfrac{\alpha+\beta}{2} = 180^\circ -\dfrac{180^\circ -\gamma}{2} & \Leftrightarrow~ \gl{3}~ y+z = 90^\circ +\dfrac\gamma2 \\[1em] \end{array}$ $\gl{2} -\gl{3}:~ x-y =\dfrac\beta2 -\dfrac\gamma2 ~~\gl{4}$ $\gl{1} ~\Leftrightarrow~ y=90^\circ +\dfrac\alpha2 -x$ in $\gl{4}$: $\begin{array}{l l l} x -90^\circ -\dfrac\alpha2 +x =\dfrac\beta2 -\dfrac\gamma2 & ~\Leftrightarrow~ 2x &=90^\circ +\dfrac{\alpha+\beta-\gamma}{2} =\dfrac{180^\circ -\gamma +\alpha +\beta}{2} \\[1em] & &=\dfrac{2\alpha +2\beta}{2} =\alpha +\beta \\ ~\Leftrightarrow~ x = \dfrac{\alpha +\beta}{2} ~~\gl{5} \\[1em] \end{array}$ $\gl{5} ~\Leftrightarrow~ x = \dfrac{180^\circ-\gamma}{2} ~\Leftrightarrow~ \underline{ x=: \dfrac{\gamma'}{2} }$ $\gl{5}$ in $\gl{1}$: $ \dfrac{\alpha +\beta}{2} +y = 90^\circ +\dfrac\alpha2$ $~\Leftrightarrow~ y =\dfrac{180^\circ +\alpha -\alpha -\beta}{2} =\dfrac{180^\circ -\beta}{2} ~\Leftrightarrow~ \underline{ y=: \dfrac{\beta'}{2} }$ $\gl{5}$ in $\gl{2}$: $ \dfrac{\alpha +\beta}{2} +z = 90^\circ +\dfrac\beta2$ $~\Leftrightarrow~ z =\dfrac{180^\circ +\beta -\alpha -\beta}{2} =\dfrac{180^\circ -\alpha}{2} ~\Leftrightarrow~ \underline{ z=: \dfrac{\alpha'}{2} }$