Ceva定理：对于三角形ABC与任意点P,PA,PB,PC交BC,CA,AB于S,Q,R.有AQ/QC*CS/SB*BR/RA=1

证明：AQ/QC=S(AQP)/S(CQP)=(AP*PQ*Sin(APQ))/(PC*PQ*Sin(CPQ))=AP/PC*Sin(APQ)/Sin(CPQ)
同理CS/SB=CP/PB*Sin(CPS)/Sin(BPS)
BR/RA=……
三式相乘即得

注意到AQ/QC=S(ABQ)/S(CBQ)=(AB*BQ*Sin(ABQ))/(BC*BQ*Sin(CBQ))
CS/SB=……BR/RA=……三式相乘，结合Ceva定理得
Sin(ABQ)/Sin(CBQ)*Sin(CAS)/Sin(BAS)*Sin(CBQ)/Sin(ABQ)=1
上式即角元Ceva定理