Wykonajmy rysunek pomocniczy:



Obliczmy pole trójkąta ABC:     

$P_{ABC}=\frac{6^2\sqrt{3}}{4}=\frac{36\sqrt{3}}{4}=9\sqrt{3}$  


I przypadek:

$\sin {75}^{\circ }=\sin \left({30}^{\circ }+{45}^{\circ }\right)=\sin {30}^{\circ }\cdot \cos {45}^{\circ }+\cos {30}^{\circ }\cdot \sin {45}^{\circ }=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}=\frac{\sqrt{2}+\sqrt{6}}{4}$  

$\frac{\left|DE\right|}{\sin {60}^{\circ }=}\frac{2}{\sin {75}^{\circ }}$  

$\frac{\left|DE\right|}{\frac{\sqrt{3}}{2}}=\frac{2}{\frac{\sqrt{2}+\sqrt{6}}{4}}$  

$2\frac{\left|DE\right|}{\sqrt{3}}=\frac{8}{\sqrt{2}+\sqrt{6}}\mid :2$  

$\frac{\left|DE\right|}{\sqrt{3}}=\frac{4}{\sqrt{2}+\sqrt{6}}$  

$\left|DE\right|=\frac{4\sqrt{3}\cdot \left(\sqrt{2}-\sqrt{6}\right)}{2-6}$  

$\left|DE\right|=\frac{4\sqrt{3}\cdot \left(\sqrt{2}-\sqrt{6}\right)}{-4}$  

$\left|DE\right|=\sqrt{18}-\sqrt{6}$  

$\left|DE\right|=3\sqrt{2}-\sqrt{6}$  


Obliczmy pole powierzchni trójkąta DBE:

$P_{DBE}=\frac{1}{2}\cdot 2\cdot \left(3\sqrt{2}-\sqrt{6}\right)\cdot \sin {45}^{\circ }=\left(3\sqrt{2}-\sqrt{6}\right)\cdot \frac{\sqrt{2}}{2}=\frac{6-\sqrt{12}}{2}=\frac{6-2\sqrt{3}}{2}=3-\sqrt{3}$  


Obliczmy stosunek pola trójkąta DBE do pola trójkąta ABC:

$\frac{P_{DBE}}{P_{ABC}}=\frac{3-\sqrt{3}}{9\sqrt{3}}=\frac{3\sqrt{3}-3}{9\cdot 3}=\frac{\sqrt{3}-1}{9}$  

---

II przypadek:

$\sin {75}^{\circ }=\sin \left({30}^{\circ }+{45}^{\circ }\right)=\sin {30}^{\circ }\cdot \cos {45}^{\circ }+\cos {30}^{\circ }\cdot \sin {45}^{\circ }=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}=\frac{\sqrt{2}+\sqrt{6}}{4}$  

$\frac{\left|DE\right|}{\sin {60}^{\circ }=}\frac{4}{\sin {75}^{\circ }}$  

$\frac{\left|DE\right|}{\frac{\sqrt{3}}{2}}=\frac{4}{\frac{\sqrt{2}+\sqrt{6}}{4}}$  

$2\frac{\left|DE\right|}{\sqrt{3}}=\frac{16}{\sqrt{2}+\sqrt{6}}\mid :2$  

$\frac{\left|DE\right|}{\sqrt{3}}=\frac{8}{\sqrt{2}+\sqrt{6}}\mid \sqrt{3}$  

$\left|DE\right|=\frac{8\sqrt{3}\cdot \left(\sqrt{2}-\sqrt{6}\right)}{2-6}$  

$\left|DE\right|=\frac{8\sqrt{3}\cdot \left(\sqrt{2}-\sqrt{6}\right)}{-4}$  

$\left|DE\right|=2\sqrt{18}-2\sqrt{6}$  

$\left|DE\right|=6\sqrt{2}-2\sqrt{6}$  


Obliczmy pole powierzchni trójkąta DBE:

$P_{DBE}=\frac{1}{2}\cdot 4\cdot \left(6\sqrt{2}-2\sqrt{6}\right)\cdot \sin {45}^{\circ }=2\cdot \left(6\sqrt{2}-2\sqrt{6}\right)\cdot \frac{\sqrt{2}}{2}=\left(6\sqrt{2}-2\sqrt{6}\right)\cdot \sqrt{2}=12-2\sqrt{12}=12-4\sqrt{3}$  


Obliczmy stosunek pola trójkąta DBE do pola trójkąta ABC:

$\frac{P_{DBE}}{P_{ABC}}=\frac{12-4\sqrt{3}}{9\sqrt{3}}=\frac{12\sqrt{3}-12}{9\cdot 3}=\frac{3\cdot \left(4\sqrt{3}-4\right)}{9\cdot 3}=\frac{4\sqrt{3}-4}{9}$