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\( \left\|\begin{array}{c}
\triangle\ ABC\\\\
\ = = = = = = = = = = = = = = = = =\ \\\\
\ D\in BC\ \ ,\ \ E\in CA\ \ ,\ \ F\in AB\ \\\\
\ P\in AD\cap CF\ \ ,\ \ R\in AD\cap BE\ \\\\
\ \frac {\overline {AE}}{\overline {EC}}=m\ \ ,\ \ \frac {\overline {AF}}{\overline {FB}}=n\ \end{array}\right\|\ \Longrightarrow\ \frac {m}{\overline {AR}} + \frac {n}{\overline {AP}}=\frac {1+m+n}{\overline {AD}} \) .

Caz particular : \( m=n=1\ \Longrightarrow\ \frac {1}{\overline {AR}} + \frac {1}{\overline {AP}}=\frac {3}{\overline {AD}} \) .

Aplicam teorema lui Menelaus in triunghiul ABD cu transversala (F,C,P)

\( \frac{BC}{CD}\cdot\frac{DP}{PA}\cdot n=1 \) de unde \( \frac{CD}{BC}=n\frac{DP}{PA} \)

Analog in triunghiul ACD cu tranversala (E,B,R). \( \frac{BC}{BD}\cdot\frac{DR}{RA}\cdot m=1 \) deci \( \frac{BD}{BC}=m\frac{DR}{RA} \)

Din \( \frac{BD}{BC}+\frac{DC}{BC}=1 \) avem \( m\frac{DR}{AR}+n\frac{DP}{AP}=1 \) relatie care este echivalenta cu cea din concluzie.