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Posted: **Mon Feb 02, 2009 1:03 pm**

\( \mathrm I\ \odot\ \ \)Fie \( \triangle\ ABC \) si punctele \( M\in (AB) \) , \( N\in (AC) \) , \( P\in BN\cap CM \) .

Sa se arate ca \( \overline {\underline {\left\|\ \begin{array}{cccc}
1. & [MPN]\cdot [BPC] & = & [BPM]\cdot [CPN]\\\\
2. & [MAN]\cdot [BPC] & = & [ABC]\cdot [MPN]\\\\
3. & [AMPN]=[BPC] & \Longleftrightarrow & \frac {MA}{MB}\cdot\frac {NA}{NC}=1\end{array}\ \right\|}} \)

Posted: **Mon Aug 10, 2009 10:22 pm**

\( 1. \) Avem \( \frac{[MPN]}{[CPN]}=\frac{[BPM]}{[BPC]}=\frac{MP}{PC}\ \Longrightarrow\ \[MPN]\cdot [BPC]=[BPM]\cdot [CPN] \) .

\( 2. \) Avem \( \left\|\begin{array}{cc}
\frac{PM}{PC} & = & \frac{AM}{AC}\ \cdot\ \frac{\sin(\angle MAP)}{\sin(\angle PAC)}\\\\\\\\\\\\
\frac{PN}{PB} & = & \frac{AN}{AB}\ \cdot\ \frac{\sin(\angle PAN)}{\sin(\angle PAB)} \end{array}\right| \bigodot\ \Longrightarrow\ \frac{PM\cdot PN}{PC\cdot PB}=\frac{AM\cdot AN}{AB\cdot AC}\ \Longleftrightarrow\ \frac{[MPN]}{[BPC]}=\frac{[MAN]}{[ABC]}
\) .

\( 3. \) Avem \( [ABC]=[AMPN]+[BMC]+[BNC]-[BPC]\ \Longleftrightarrow\ [AMPN]-[BPC]=[ABC]- \)

\( -[BMC]-[BNC]\ \Longleftrightarrow\ [AMPN]-[BPC]=[ABC]\left\(1-\frac{MB}{AB}-\frac{NC}{AC}\right\) \)

\( \Longrightarrow\ [AMPN]=[BPC]\ \Longleftrightarrow\ \frac{MB}{AB}+\frac{NC}{AC}=1\ \Longleftrightarrow\ \frac{MB}{MA}=\frac{NA}{NC}\ \Longleftrightarrow\ \frac{MA}{MB}\cdot \frac{NA}{NC}=1 \) .

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