考虑等角六边形 $ABCDEF$，求证：\[AC^2+CE^2+EA^2=BD^2+DF^2+FB^2.\]
证明
等角六边形 $ABCDEF$ 的六个内角均为 $120^\circ$，根据余弦定理，有\[AC^2=AB^2+BC^2+AB\cdot BC,\]类似的计算其他对角线的平方，可得欲证等式即\[AB\cdot BC+CD\cdot DE+EF\cdot FA=BC\cdot CD+DE\cdot EF+FA\cdot AB,\]也即\[[ABC]+[CDE]+[EFA]=[BCD]+[DEF]+[FAB],\]也即\[[ACE]=[BDF].\]
作平行六边形 $ABCDEF$ 的伴随三角形 $PQR$，设 $\dfrac{PE}{EQ}=\lambda_1$，$\dfrac{QA}{AR}=\lambda_2$，$\dfrac{RC}{CP}=\lambda_3$，因为 $DE\parallel QR$，$FA\parallel RP$，$BC\parallel PQ$，所以\[\dfrac{PD}{DR}=\dfrac{PE}{EQ}=\lambda_1,\quad \dfrac{QF}{FP}=\dfrac{QA}{AR}=\lambda_2,\quad \dfrac{RB}{BQ}=\dfrac{RC}{CP}=\lambda_3,\]所以\[\begin{split}\dfrac{[PCE]}{[PQR]}&=\dfrac{PC\cdot PE}{PQ\cdot PR}\\ &=\dfrac{PC\cdot PE}{(PE+EQ)(PC+CR)}\\ &=\dfrac{1}{\left(1+\dfrac{EQ}{PE}\right)\left(1+\dfrac{RC}{CP}\right)}\\ &=\dfrac{\lambda_1}{(1+\lambda_1)(1+\lambda_3)},\end{split}\]同理可得\[\dfrac{[QEA]}{[PQR]}=\dfrac{\lambda_2}{(1+\lambda_2)(1+\lambda_3)},\quad \dfrac{[RAC]}{[PQR]}=\dfrac{\lambda_3}{(1+\lambda_3)(1+\lambda_2)},\]从而\[\dfrac{[ACE]}{[PQR]}=\dfrac{1+\lambda_1\lambda_2\lambda_3}{(1+\lambda_1)(1+\lambda_2)(1+\lambda+3)},\]同理可得\[\dfrac{[BDF]}{[PQR]}=\dfrac{1+\lambda_1\lambda_2\lambda_3}{(1+\lambda_1)(1+\lambda_2)(1+\lambda+3)},\]因此\[[ACE]=[BDF],\]命题得证．