$$\dfrac{s_1}{s}=\dfrac{s_1}{\left[ABD\right]}\cdot \dfrac{\left[ABD\right]}{s}=\dfrac{AN\cdot AK}{AD\cdot AB}\cdot \dfrac{\left[ABD\right]}{s}=\dfrac{AN}{AD}\cdot \dfrac{AK}{AB}\cdot \dfrac{\left[ABD\right]}{s}$$ $$\Rightarrow \sqrt[3]{\dfrac{s_1}{s}}=\sqrt[3]{\dfrac{AN}{AD}\cdot \dfrac{AK}{AB}\cdot \dfrac{\left[ABD\right]}{s}{\rm \ }}.$$ $AO\ge GO$ dan $\dfrac{\dfrac{AN}{AD}+\dfrac{AK}{AB}+\dfrac{\left[ABD\right]}{s}}{3}\ge \sqrt[3]{\dfrac{AN}{AD}\cdot \dfrac{AK}{AB}\cdot \dfrac{\left[ABD\right]}{s}{\rm \ }}=\sqrt[3]{\dfrac{s_1}{s}}$ elde edilir.
Benzer şekilde $$\dfrac{\dfrac{BK}{AB}+\dfrac{BL}{BC}+\dfrac{\left[BAC\right]}{s}}{3}\ge \sqrt[3]{\dfrac{s_2}{s}}, $$ $$\dfrac{\dfrac{CL}{BC}+\dfrac{CM}{CD}+\dfrac{\left[CBD\right]}{s}}{3}\ge \sqrt[3]{\dfrac{s_3}{s}},$$ $$\dfrac{\dfrac{DM}{CD}+\dfrac{DN}{AD}+\dfrac{\left[DAC\right]}{s}}{3}\ge \sqrt[3]{\dfrac{s_4}{s}}$$ elde edilir. Taraf tarafa toplarsak, $$\frac{\frac{AN}{AD}+\frac{DN}{AD}+\frac{AK}{AB}+\frac{BK}{AB}+\frac{BL}{BC}+\frac{CL}{BC}+\frac{CM}{CD}+\frac{DM}{CD}+\frac{\left[ABD\right]+\left[BCD\right]+\left[BAC\right]+\left[DAC\right]}{s}}{3}$$ $$\ge \dfrac{\sqrt[3]{s_1}+\sqrt[3]{s_2}+\sqrt[3]{s_3}+\sqrt[3]{s_4}}{\sqrt[3]{s}}$$ olur. Düzenlersek, $$\dfrac{1+1+1+1+\dfrac{s+s}{s}}{3}=\dfrac{6}{3}=2\ge \dfrac{\sqrt[3]{s_1}+\sqrt[3]{s_2}+\sqrt[3]{s_3}+\sqrt[3]{s_4}}{\sqrt[3]{s}}$$ Eşitlik durumu,
$$\dfrac{AN}{AD}=\dfrac{AK}{AB}=\dfrac{\left[ABD\right]}{s}\Rightarrow \dfrac{AN}{DN}=\dfrac{AK}{BK}=\dfrac{\left[ABD\right]}{\left[CBD\right]},$$ $$\dfrac{BK}{AB}=\dfrac{BL}{BC}=\dfrac{\left[BAC\right]}{s}\Rightarrow \dfrac{BK}{AK}=\dfrac{CL}{BL}=\dfrac{\left[BAC\right]}{\left[CAD\right]},$$ $$\dfrac{CL}{BC}=\dfrac{CM}{CD}=\dfrac{\left[CBD\right]}{s}\Rightarrow \dfrac{CL}{BL}=\dfrac{CM}{DM}=\dfrac{\left[CBD\right]}{\left[ABD\right]},$$ $$\dfrac{DM}{CD}=\dfrac{DN}{AD}=\dfrac{\left[DAC\right]}{s}\Rightarrow \dfrac{DM}{CM}=\dfrac{DN}{AN}=\dfrac{\left[DAC\right]}{\left[ABC\right]},$$ iken sağlanır. Birleştirsek, $KLMN$ kenarları $ABCD$ nin köşegenlere paralel olan bir paralelkenar ve $$\dfrac{AK}{BK}=\dfrac{\left[ABD\right]}{\left[CBD\right]}=\dfrac{\left[CAD\right]}{\left[BAC\right]}=\dfrac{\left[CBD\right]}{\left[ABD\right]}=\dfrac{\left[BAC\right]}{\left[CAD\right]}$$ olur. Bu durumda $\left[ABD\right]=\left[CBD\right]$ ve $\left[CAD\right]=\left[BAC\right]$ olduğu için köşegenler birbirlerini ortalar, yani $ABCD$ paralelkenar olur.
Yani, eşitlik durumu $ABCD$ dörtgeni paralelkenarken ve $K,L,M,N$ noktaları orta noktalar iken sağlanır.
