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JBTST V 2010, Problema 2
Posted: Sun May 23, 2010 12:03 pm
by Andi Brojbeanu
FF simpla !
Fie \( ABC \) un triunghi si \( D, E, F \) mijloacele laturilor \( BC, CA, AB \) respectiv.
Sa se arate ca \( \angle{DAC}=\angle{ABE} \) daca si numai daca \( \angle{AFC}=\angle{BDA} \) .
Posted: Sun May 23, 2010 5:18 pm
by Claudiu Mindrila
Intai \( DF\parallel AC\Longrightarrow\widehat{ADF}=\widehat{DAC} \). Fie \( G \) centrul de greutate al triunghiului \( ABC \).
Acum,
\( \widehat{DAC}=\widehat{ABE}\Longleftrightarrow\widehat{ABE}=\widehat{ADF}\Longleftrightarrow\widehat{FBG}=\widehat{FDG}\Longleftrightarrow BDFG\ \text{inscriptibil}\Longleftrightarrow\widehat{AFC}=\widehat{BDA} \)
Posted: Mon May 24, 2010 5:28 am
by Virgil Nicula
Fie \( ABC \) un triunghi si \( D, E, F \) mijloacele laturilor \( BC, CA, AB \) respectiv.
Sa se arate ca \( \angle{DAC}=\angle{ABE} \) daca si numai daca \( \angle{AFC}=\angle{BDA} \) .
Observatii.
1 - \( \ \widehat{DAC}\equiv\widehat{ABE}\ \Longleftrightarrow\ \widehat{AFC}\equiv\widehat{BDA}\ \Longleftrightarrow\ a^2+c^2=2b^2 \) .
2 - Fie
\( ABC \) un triunghi si mijlocul
\( E \) al laturii
\( CA \) . Pentru un punct
\( M\in (BE) \) notam
\( D\in AM\cap BC \) si
\( F\in CM\cap AB \) .
Sa se arate ca
\( \angle{DAC}=\angle{ABE}\ \Longleftrightarrow\ \angle{AFC}=\angle{BDA}\ \Longleftrightarrow\ (1-2m)(a^2+c^2)=(1-m)b^2 \) , unde
\( MD=m\cdot AD \) .