Fie triunghiul ABC si EF paralela cu BC, E apartine (AB), F apartine (AC). Sa se demonstreze ca AB=BC, daca si numai daca BE+EF=BC.
Stefan Smarandache,Bucuresti
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\( "=>"EF||BC \) \( secanta \) \( (EB) => \hat E \equiv \hat B \) \( (corespondete) \) 1
\( EF||BC \) \( secanta \) \( (FC) => \hat F \equiv \hat C \) \( (corespondete) \) 2
Din 1 si 2 \( => \hat A \equiv \hat E \equiv \hat F =>\triangle AEF \) echilateral
\( =>AE \equiv EF => BE + EF = BE + FA = AB => [AB]\equiv[BC] \)
\( "<=" \) \( Fie \) \( D \) \( mijlocul \) \( [BC] \) \( unim \) \( F \) \( cu \) \( D \) \( si \) \( formam \) \( \triangle CDF \) \( =>\hat D \equiv \hat B \) \( (corespondente)1 \)
\( => \)\( DF||AE \) \( secanta \) \( EF \) \( => \hat {CFD} \equiv \hat A \) \( (corespondente)2 \)
\( Din \) \( 1 \) \( si \) \( 2 \) \( => \) \( \triangle EAF \equiv \triangle DCF \) , \( [DF]\equiv[EF] \) \( => \) \( \triangle BAC \) \( echilateral \) \( => \) \( BD+DF=BC \) , \( BE+EF=AC \) \( => \) \( [AB]\equiv [BC] \)
\( EF||BC \) \( secanta \) \( (FC) => \hat F \equiv \hat C \) \( (corespondete) \) 2
Din 1 si 2 \( => \hat A \equiv \hat E \equiv \hat F =>\triangle AEF \) echilateral
\( =>AE \equiv EF => BE + EF = BE + FA = AB => [AB]\equiv[BC] \)
\( "<=" \) \( Fie \) \( D \) \( mijlocul \) \( [BC] \) \( unim \) \( F \) \( cu \) \( D \) \( si \) \( formam \) \( \triangle CDF \) \( =>\hat D \equiv \hat B \) \( (corespondente)1 \)
\( => \)\( DF||AE \) \( secanta \) \( EF \) \( => \hat {CFD} \equiv \hat A \) \( (corespondente)2 \)
\( Din \) \( 1 \) \( si \) \( 2 \) \( => \) \( \triangle EAF \equiv \triangle DCF \) , \( [DF]\equiv[EF] \) \( => \) \( \triangle BAC \) \( echilateral \) \( => \) \( BD+DF=BC \) , \( BE+EF=AC \) \( => \) \( [AB]\equiv [BC] \)
Last edited by deleter on Fri Mar 21, 2008 7:19 pm, edited 26 times in total.