Fie ABCD un patrulater convex, {M}=AB \( \cap \) CD, {N}=AD\( \cap \)BC. Sa se demonstreze ca are loc relatia:
\( \frac {MA} {MB} \cdot \frac {MC} {MD}=\frac {NA} {NB} \cdot \frac {NC} {ND} \).
Aplicatie la T.Ceva si T.Menelaus ex.6
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\( \mbox{Aplicam, mai intai, t. lui Menelaus in triunghiul ABN, relativ la transversala M-C-D: } \frac{MA}{MB}.\frac{CB}{CN}.\frac{DN}{DA}=1; \ (1)\\
\mbox{aplicam apoi aceeasi teorema la triunghiul CDN si transversala: M-A-B: } \frac{MC}{MD}.\frac{AD}{AN}.\frac{BN}{BC}=1.\ (2)\\
\mbox{Inmultind acum relatiile (1) si (2), membru cu membru, obtinem: \frac{MA}{MB}.\frac{MC}{MD}.\frac{ND}{NA}.\frac{NB}{NC}=1\Rightarrow \frac{MA}{MB}.\frac{MC}{MD}=\frac{NA}{ND}.\frac{NC}{NB}.
\)
\mbox{aplicam apoi aceeasi teorema la triunghiul CDN si transversala: M-A-B: } \frac{MC}{MD}.\frac{AD}{AN}.\frac{BN}{BC}=1.\ (2)\\
\mbox{Inmultind acum relatiile (1) si (2), membru cu membru, obtinem: \frac{MA}{MB}.\frac{MC}{MD}.\frac{ND}{NA}.\frac{NB}{NC}=1\Rightarrow \frac{MA}{MB}.\frac{MC}{MD}=\frac{NA}{ND}.\frac{NC}{NB}.
\)