Rysunek pomocniczy:

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$\text{a)}$  

$P_{ADC}=\frac{1}{2}\cdot \left|AC\right|\cdot \left|CD\right|\cdot \sin {120}^{\circ }$  

$P_{ADC}=\frac{1}{2}\cdot x\cdot \left|CD\right|\cdot \sin \left({90}^{\circ }+{30}^{\circ }\right)$      

$P_{ADC}=\frac{1}{2}\cdot x\cdot \left|CD\right|\cdot \cos {30}^{\circ }$  

$P_{ADC}=\frac{1}{2}\cdot x\cdot \left|CD\right|\cdot \frac{\sqrt{3}}{2}$  

 

$P_{DBC}=\frac{1}{2}\cdot \left|CD\right|\cdot \left|BC\right|\cdot \sin {30}^{\circ }$  

$P_{DBC}=\frac{1}{2}\cdot \left|CD\right|\cdot x\cdot \frac{1}{2}$  

 

Zatem otrzymujemy:

$\frac{P_{ADC}}{P_{DBC}}=\frac{\frac{1}{2}x\cdot \left|CD\right|\cdot \frac{\sqrt{3}}{2}}{\frac{1}{2}\cdot \left|CD\right|\cdot x\cdot \frac{1}{2}}=\frac{\sqrt{3}}{1}$  

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$\text{b)}$  

Wyznaczmy pole trójkąta $ABC$.

$P_{ABC}=\frac{1}{2}\cdot \left|AC\left|\cdot \right|BC\right|\cdot \sin \left(120^{\circ }+30^{\circ }\right)$

$P_{ABC}=\frac{1}{2}\cdot x\cdot x\cdot \sin \left(90^{\circ }+60^{\circ }\right)$

$P_{ABC}=\frac{1}{2}{x}^2\cdot \cos 60^{\circ }$

$P_{ABC}=\frac{1}{2}{x}^2\cdot \frac{1}{2}$

$P_{ABC}=\frac{1}{4}{x}^2$

Z drugiej strony pole trójkąta $ABC$ to suma pól trójkątów $ADC$ i $DBC$, zatem mamy:

$P_{ADC}+P_{DBC}=P_{ABC}$

$\sqrt{3}{P}_{DBC}+P_{DBC}=\frac{1}{4}{x}^2$

$P_{DBC}\left({\sqrt{3}}_{}+1\right)=\frac{1}{4}{x}^2$

$\frac{1}{2}\cdot \left|CD\right|\cdot \left|BC\right|\cdot \sin 30^{\circ }\left(\sqrt{3}+1\right)=\frac{1}{4}{x}^2$

$\frac{1}{2}\cdot \left|CD\right|\cdot x\cdot \frac{1}{2}\left(\sqrt{3}+1\right)=\frac{1}{4}{x}^2|:\frac{1}{4}x$

$\left|CD\right|\cdot \left(\sqrt{3}+1\right)=x|:\left|CD\right|$

$\sqrt{3}+1=\frac{x}{\left|CD\right|}$

$\sqrt{3}+1=\frac{\left|AC\right|}{\left|CD\right|}$