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如图,由梅氏定理知:\[ \frac{PE}{EB}\cdot \frac{BC}{CD}\cdot \frac{DA}{AP}=1 \]\[ \frac{PF}{FC}\cdot \frac{CB}{BD}\cdot \frac{DA}{AP}=1 \]两式相除得$ \frac{BD}{CD}=\frac{FP\cdot BE}{EP\cdot CF} $,故\[ \frac{BP\cdot BE-BF\cdot BA}{CP\cdot CF-CE\cdot CA}=\frac{BE(BP-BU)}{CF(CP-CV)}=\frac{UP\cdot BE}{VP\cdot CF}=\frac{FP\cdot BE}{EP\cdot CF}=\frac{BD}{CD} \] |
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