シムソンの定理
{\displaystyle {\begin{aligned}4((e-d)({\bar {f}}-{\bar {d}})-({\bar {e}}-{\bar {d}})(f-d))&=(({\bar {a}}b-{\bar {ab}})(|c|^{2}-1)+({\bar {b}}c-{\bar {bc}})(|a|^{2}-1)+({\bar {ac}}-{\bar {ca}})(|b|^{2}-1)\\&\ +a(|c|^{2}-|b|^{2})+b(|a|^{2}-|c|^{2})+c(|b|^{2}-|a|^{2})\\&\ +{\bar {a}}(|b|^{2}-|c|^{2})+{\bar {b}}(|c|^{2}-|a|^{2})+{\bar {c}}(|a|^{2}-|b|^{2})).\end{aligned}}}
となり、 | a | = | b | = | c | = 1 {\displaystyle |a|=|b|=|c|=1} だから ( e − d ) ( f ̄ − d ̄ ) − ( e ̄ − d ̄ ) ( f − d ) = 0 {\displaystyle (e-d)({\bar {f}}-{\bar {d}})-({\bar {e}}-{\bar {d}})(f-d)=0} となって、3点L,M,Nは同一直線上にある。Q.E.D.