Z tw. Pitagorasa dla trójkąta CDE:

${\left|DE\right|}^2+{\left|EC\right|}^2={\left|CD\right|}^2$  

$2^2+6^2={\left|CD\right|}^2$  

$4+36={\left|CD\right|}^2$  

$40={\left|CD\right|}^2$  

$\left|CD\right|=2\sqrt{10}$   

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AB||CE, więc z tw. Talesa:

$\frac{\left|BC\right|}{\left|CD\right|}=\frac{\left|AE\right|}{\left|DE\right|}$  

$\frac{\left|BC\right|}{2\sqrt{10}}=\frac{8}{2}$  

$\frac{\left|BC\right|}{2\sqrt{10}}=4\mid \cdot 2\sqrt{10}$  

$\left|BC\right|=8\sqrt{10}$  

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Z tw. Pitagorasa dla trójkąta ABD:

${\left|AD\right|}^2+{\left|AB\right|}^2={\left|BD\right|}^2$  

${\left(8+2\right)}^2+{\left|AB\right|}^2={\left(2\sqrt{10}+8\sqrt{10}\right)}^2$  

${10}^2+{\left|AB\right|}^2={\left(10\sqrt{10}\right)}^2$  

$100+{\left|AB\right|}^2=100\cdot 10$  

$100+{\left|AB\right|}^2=100$  

${\left|AB\right|}^2=900$

$\left|AB\right|=30$  

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Z tw. Pitagorasa dla trójkąta ACE:

${\left|AE\right|}^2+{\left|EC\right|}^2={\left|AC\right|}^2$  

$8^2+6^2={\left|AC\right|}^2$  

$64+36={\left|AC\right|}^2$  

$100={\left|AC\right|}^2$  

$\left|AC\right|=10$  

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Z tw. Pitagorasa dla trójkąta ABE:

${\left|AE\right|}^2+{\left|AB\right|}^2={\left|BE\right|}^2$

$8^2+{30}^2={\left|BE\right|}^2$  

$64+900={\left|BE\right|}^2$  

$964={\left|BE\right|}^2$  

$\left|BE\right|=2\sqrt{241}$  

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$\text{Odp.}\left|AC\right|=10,\left|BE\right|=2\sqrt{241}.$