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In tetraedrul \( ABCD \) punctele \( E \) si \( F \) sunt mijloacele medianelor \( AM \) si \( AN \) ale triunghiurilor \( ABC \) respectiv \( ACD \). Daca \( CE\cap AB={P} \),\( CF\cap AD={Q} \), \( DF\cap AC={R} \), demonstrati ca:
a)\( 9 \)Aria\( (PQR) \)=Aria\( (BDC) \);
b)\( 12(PQ+EF+MN)=13BD \).

Virginia si Vasile Tica, Campulung

a) Evident, punctele \( P, Q, R \) sunt mijloacele muchiilor \( [AB], [AD], [AC] \).
Din reciproca teoremei lui Menelaus in \( \bigtriangleup{AMC} \) si secanta \(
B-E-R \) obtinem: \( \frac\frac{AR}{RC}=\frac{MB}{BC}\cdot \frac{AE}{EM}=\frac{1}{2} \). Folosind relatia lui Van Aubel, avem:\( \frac{AP}{PB}=\frac{AE}{EM}-\frac{AR}{RC}=1-\frac{1}{2}=\frac{1}{2}=\frac{AR}{RC}\Rightarrow PR\parallel BC \) si \( PR=\frac{AR}{AR+RC}\cdot BC=\frac{1}{3}\cdot BC \).
Analog, \( QR\parallel CD \) si \( QR=\frac{1}{3}\cdot CD \).
Deoarece \( PR\parallel BC \) si \( QR\parallel CD\Rightarrow \angle{PQR}\equiv \angle{BCD} \).
Atunci \( S_{BCD}=\frac{BC\cdot CD\cdot sin(\angle{BCD})}{2}=\frac{3PQ\cdot 3QR\cdot sin (\angle{BCD})}{2}=9\cdot \frac{PQ\cdot QR\cdot sin (\angle{PQR})}{2}=9S_{PQR} \).
b)Deoarece \( [MN] \)si \( [EF] \) sunt liniii mijlocii in \( \bigtriangleup{BCD} \) si \( \bigtriangleup{AMN}\Rightarrow MN=\frac{1}{2}\cdot BD \) si \( EF=\frac{1}{2}\cdot MN=\frac{1}{4}\cdot BD \). De asemenea, cum \( \frac{AP}{PB}=\frac{AQ}{QD}=\frac{1}{2}\Rightarrow PQ\parallel BD \) si apoi \( PQ=\frac{1}{3}\cdot BD \)
In final, \( 12(PQ+EF+MN)=12\cdot BD(\frac{1}{2}+\frac{1}{3}+\frac{1}{4})=12BD\cdot \frac{6+4+3}{12}=13BD \).