0.0.1 ↑ 76. Hausaufgabe
0.0.1.1 ↑ Geometrie-Buch Seite 117, Aufgabe 1
Im Dreieck ABC ist \overrightarrow{AD} = \frac{2}{3} \overrightarrow{AC} und \overrightarrow{BE} = \frac{3}{5} \overrightarrow{BC}.
In welchen Verhältnissen teilen sich \left[AE\right] und \left[BD\right]?
\overrightarrow{AS} = \alpha \overrightarrow{SE}; \quad {}\overrightarrow{BS} = \beta \overrightarrow{SD};
\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;
\lambda + \frac{1}{3} \mu - 1 = -\frac{2}{5} - \mu + 1 = 0; ⇔ (\lambda,\mu) = (\frac{4}{5},\frac{3}{5});
\overrightarrow{AS} = \alpha \overrightarrow{SE} = \lambda \overrightarrow{AE} = \lambda \left(\overrightarrow{AS} + \overrightarrow{SE}\right); ⇔ \alpha = \lambda \frac{\overrightarrow{AS}}{\overrightarrow{SE}} + \lambda = \lambda \alpha + \lambda; ⇔ \alpha = \frac{\lambda}{1 - \lambda} = 4;
\overrightarrow{BS} = \beta \overrightarrow{SD} = -\mu \overrightarrow{DB} = -\mu \left(\overrightarrow{DS} + \overrightarrow{SB}\right) = \mu \overrightarrow{SD} + \mu \overrightarrow{BS}; ⇔ \beta = \mu \frac{\overrightarrow{SD}}{\overrightarrow{SD}} + \mu \frac{\overrightarrow{BS}}{\overrightarrow{SD}} = \mu + \mu \beta; ⇔ \beta = \frac{\mu}{1 - \mu} = \frac{3}{2};
[XXX falsch.]