\ANG{
Das Dreieck ABC ist rechtwinklig mit einem rechten Winkel bei $C$. Der
Fu"spunkt $D$ der H"ohe $h$ zerlegt $[AB]$ in zwei Abschnitte der L"angen
$p$ und $q$.

\begin{minipage}[t]{7cm}
\begin{enumerate}
\item
Zeige: Die Dreiecke $ADC$ und $CDB$ sind "ahnlich.
\item
Folgere daraus: $h^2 = pq$.
\end{enumerate}
\end{minipage}
\hfill 
\begin{minipage}[t]{7cm}
\beginpicture
\setcoordinatesystem units <0.85mm,0.85mm> point at 0 35
\unitlength0.85mm
\setplotarea x from -3 to 78, y from -5 to 40
\plot 27 36 0 0 75 0 27 36 27 0 /
\put {$A$} [tr] <-1mm,-1mm> at 0 0
\put {$B$} [tl] <1mm,-1mm> at 75 0
\put {$C$} [B] <0mm,1mm> at 27 36
\put {$D$} [t] <0mm,-1mm> at 27 0
\circulararc 180 degrees from 30 0 center at 27 0
\put {\circle*{1}} [Bl] <-1.2mm,1.2mm> at 27 0
\put {\circle*{1}} [Bl] <1.2mm,1.2mm> at 27 0
\put {$\gamma_1$} [tr] <-0.1mm,-6mm> at 27 36
\circulararc -36.870 degrees from 27 26 center at 27 36
\put {$\gamma_2$} [tl] <1.2mm,-5mm> at 27 36
\circulararc 53.130 degrees from 27 26 center at 27 36
\put {$\alpha$} [Bl] <5mm,2mm> at 0 0
\circulararc 53.130 degrees from 10 0 center at 0 0
\put {$\beta$} [Br] <-6mm,1mm> at 75 0
\circulararc -36.870 degrees from 65 0 center at 75 0
\put {$h$} [Bl] <1mm,0mm> at 27 18
\put {$p$} [t] <0mm,-1mm> at 13.5 0
\put {$q$} [t] <0mm,-1mm> at 51 0
\endpicture
\end{minipage}
}
\LSG{\mbox{}
}
