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Di Mai  9 20:15:13 CEST 2006

=head2 77. Hausaufgabe

=head3 Geometrie-Buch Seite 117, Aufgabe 2

Im Dreieck M<ABC> ist M<\overrightarrow{BD} = \frac{3}{4} \overrightarrow{BD}> und M<\overrightarrow{AS} = \frac{1}{2}
\overrightarrow{AD}>. M<BS> schneidet M<AC> in M<T>.

In welchem Verhältnis teilt M<T> die Strecke M<\overrightarrow{AC}> beziehungsweise M<S> die
Strecke M<\overrightarrow{BT}>?

M<\overrightarrow{AB}>, M<\overrightarrow{AC}> linear unabhängig.

M<\overrightarrow{AS} + \overrightarrow{ST} + \overrightarrow{TA} =\\{\quad}= \frac{1}{2} \underbrace{\left(\overrightarrow{AB} + \frac{3}{4}
\overrightarrow{BC}\right)}_{\overrightarrow{AD}} + \underbrace{\lambda \overrightarrow{BT}}_{\overrightarrow{ST}} + \underbrace{\mu
\overrightarrow{CA}}_{\overrightarrow{TA}} =\\{\quad}= \frac{1}{2} \overrightarrow{AB} + \frac{3}{8} \underbrace{\left(\overrightarrow{BA} +
\overrightarrow{AC}\right)}_{\overrightarrow{BC}} + \lambda \underbrace{\left(\overrightarrow{BA} - \mu \overrightarrow{CA}\right)}_{\overrightarrow{BT}}
+ \mu \overrightarrow{CA} =\\{\quad}= \overrightarrow{AB} \left(\frac{1}{2} - \frac{3}{8} - \lambda\right) + \overrightarrow{AC}
\left(\frac{3}{8} + \lambda\mu - \mu\right) =\\{\quad}= \vec 0;>

M<\lambda = \frac{1}{2} - \frac{3}{8} = \frac{1}{8};>

M<\mu = \frac{\frac{3}{8}}{1 - \lambda} = \frac{3}{7};>

M<\overrightarrow{BS} = \beta \overrightarrow{ST};> ⇔ M<\beta = \frac{\overrightarrow{BS}}{\overrightarrow{ST}} = \frac{\overrightarrow{BT} +
\overrightarrow{TS}}{\lambda \overrightarrow{BT}} = \frac{\overrightarrow{BT} - \lambda \overrightarrow{BT}}{\lambda \overrightarrow{BT}} = \frac{1 -
\lambda}{\lambda} = 7;>

M<\overrightarrow{AT} = \alpha \overrightarrow{TC};> ⇔ M<\alpha = \frac{\overrightarrow{AT}}{\overrightarrow{TC}} = \frac{\overrightarrow{AT}}{\overrightarrow{TA} +
\overrightarrow{AC}} = \frac{\mu \overrightarrow{AC}}{-\mu \overrightarrow{AC} + \overrightarrow{AC}} = \frac{\mu}{1 - \mu} = \frac{3}{4};>