=for timestamp
Fr Mai  5 15:49:05 CEST 2006

=head2 76. Hausaufgabe

=head3 Geometrie-Buch Seite 117, Aufgabe 1

Im Dreieck M<ABC> ist M<\overrightarrow{AD} = \frac{2}{3} \overrightarrow{AC}>
und M<\overrightarrow{BE} = \frac{3}{5} \overrightarrow{BC}>.

In welchen Verhältnissen teilen sich M<\left[AE\right]> und M<\left[BD\right]>?

M<\overrightarrow{AS} = \alpha \overrightarrow{SE}; \quad
{}\overrightarrow{BS} = \beta  \overrightarrow{SD};>

M<\overrightarrow{AS} + \overrightarrow{SB} + \overrightarrow{BA} = \underbrace{\lambda \overrightarrow{AE}}_{\overrightarrow{AS}} + \underbrace{\mu \overrightarrow{DB}}_{\overrightarrow{SB}} + \underbrace{\overrightarrow{BC} +
\overrightarrow{CA}}_{\overrightarrow{BA}} = \underbrace{\lambda \left(\overrightarrow{AC} + \frac{2}{5} \overrightarrow{CB}\right)}_{\overrightarrow{AS}} + \underbrace{\mu
\left(\frac{1}{3} \overrightarrow{AC} + \overrightarrow{CB}\right)}_{\overrightarrow{SB}} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{AC} \left(\lambda
+ \frac{1}{3} \mu - 1\right) + \overrightarrow{BC} \left(-\frac{2}{5} - \mu + 1\right) = \vec
0;>

M<\lambda + \frac{1}{3} \mu - 1 = -\frac{2}{5} - \mu + 1 = 0;> ⇔
M<(\lambda,\mu) = (\frac{4}{5},\frac{3}{5});>

M<\overrightarrow{AS} = \alpha \overrightarrow{SE} = \lambda \overrightarrow{AE} = \lambda \left(\overrightarrow{AS} + \overrightarrow{SE}\right);> ⇔
M<\alpha = \lambda \frac{\overrightarrow{AS}}{\overrightarrow{SE}} + \lambda = \lambda \alpha + \lambda;> ⇔
M<\alpha = \frac{\lambda}{1 - \lambda} = 4;>

M<\overrightarrow{BS} = \beta  \overrightarrow{SD} = -\mu    \overrightarrow{DB} = -\mu    \left(\overrightarrow{DS} + \overrightarrow{SB}\right) = \mu
\overrightarrow{SD} + \mu \overrightarrow{BS};> ⇔
M<\beta = \mu \frac{\overrightarrow{SD}}{\overrightarrow{SD}} + \mu \frac{\overrightarrow{BS}}{\overrightarrow{SD}} = \mu + \mu \beta;> ⇔
M<\beta = \frac{\mu}{1 - \mu} = \frac{3}{2};>

[XXX falsch.]