$I\mathrm{.}{\displaystyle \frac{a+b+c}{i}}={\displaystyle \frac{a}{g}}I.\backslash quad\backslash dfrac\{a+b+c\}\{i\}=\backslash dfrac\{a\}\{g\}$ (2. Strahlensatz)

$II\mathrm{.}{\displaystyle \frac{c}{f}}={\displaystyle \frac{b}{e}}II.\backslash hspace\{14mm\}\backslash dfrac\{c\}\{f\}=\backslash dfrac\{b\}\{e\}$ (1. Strahlensatz)

$III\mathrm{.}{\displaystyle \frac{d+e}{h}}={\displaystyle \frac{d}{g}}III.\backslash hspace\{6mm\}\backslash dfrac\{d+e\}\{h\}=\backslash dfrac\{d\}\{g\}$ (2. Strahlensatz)

Es gilt: ${\displaystyle \frac{a+b}{h}}={\displaystyle \frac{a}{g}}\Rightarrow {\displaystyle \frac{(4+6)cm}{h}}={\displaystyle \frac{4cm}{2cm}}\backslash quad\backslash dfrac\{a+b\}\{h\}=\backslash dfrac\{a\}\{g\}\backslash \; \backslash Rightarrow\backslash dfrac\{\; (4+6)\backslash \; cm\}\{h\}=\backslash dfrac\{4\backslash \; cm\}\{2\backslash \; cm\}\backslash qquad$(2. Strahlensatz)

$4h=20cm\Rightarrow h=5cm\backslash hspace\{19mm\}4h=20\backslash \; cm\backslash \; \backslash Rightarrow\backslash \; h=5\backslash \; cm$

Die Länge der Strecke$h\backslash \; h\backslash$beträgt$\boxed{{\displaystyle 5cm}}\backslash \; \backslash boxed\{5\backslash \; cm\}$