Rysunek pomocniczy:

   

Korzystając z tw. sinusów dla trójkąta ADE mamy:

$\frac{\left|AD\right|}{\sin \left({180}^{\circ }-\left({75}^{\circ }+{45}^{\circ }\right)\right)}=\frac{\left|ED\right|}{\sin {75}^{\circ }}$  

$\frac{\frac{a\sqrt{3}}{2}}{\sin {60}^{\circ }}=\frac{\left|ED\right|}{\sin \left({45}^{\circ }+{30}^{\circ }\right)}$  

$\frac{\frac{a\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}=\frac{\left|ED\right|}{\sin {45}^{\circ }\cos {30}^{\circ }+\cos {45}^{\circ }\sin {30}^{\circ }}$  

$a=\frac{\left|ED\right|}{\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot \frac{1}{2}}$  

$a\cdot \left(\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}\right)=\left|ED\right|$  

$\frac{a\left(\sqrt{6}+\sqrt{2}\right)}{4}=\left|ED\right|$  

 

Wyznaczmy pole tego przekroju.

$P_{BCE}=\frac{1}{2}\cdot \left|BC\right|\cdot \left|ED\right|$  

$P_{BCE}=\frac{1}{2}\cdot a\cdot \frac{a\left(\sqrt{6}+\sqrt{2}\right)}{4}$  

$P_{BCE}=\frac{a^2\left(\sqrt{6}+\sqrt{2}\right)}{8}$