Fie patrulaterul convex \( ABCD \) pentru care notam \( O\in AC\cap BD \) . Bisectoarea unghiului \( \angle AOB \)
intalneste \( AB \) , \( CD \) in punctele \( E \) , \( F \) respectiv. Sa se arate ca \( \frac {OE}{OF}\ =\ \frac {OA\cdot OB}{OA+OB}\ \cdot\ \frac {OC+OD}{OC\cdot OD} \) .
O.L.M. Bucuresti (geometrie).
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Mateescu Constantin
- Newton
- Posts: 307
- Joined: Tue Apr 21, 2009 8:17 am
- Location: Pitesti
Scriem lungimile bisectoarelor \( OE \) si \( OF \) in triunghiurile \( AOB \) si respectiv \( COD\ : \)
\( \left\|\ \begin{array}{cc}
OE=\frac{2\cdot OA\cdot OB}{OA+OB} & \cdot & \cos\frac{\widehat{AOB}}{2} \\\\\\\\
OF=\frac{2\cdot OC\cdot OD}{OC+OD} & \cdot & \cos\frac{\widehat{COD}}{2}\ \end{array}\right|\ (\div)\ \Longrightarrow\ \frac{OE}{OF}=\frac{OA\cdot OB}{OA+OB}\ \cdot\ \frac{OC+OD}{OC\cdot OD} \) , intrucat \( \angle AOB\equiv\angle COD \) .
\( \left\|\ \begin{array}{cc}
OE=\frac{2\cdot OA\cdot OB}{OA+OB} & \cdot & \cos\frac{\widehat{AOB}}{2} \\\\\\\\
OF=\frac{2\cdot OC\cdot OD}{OC+OD} & \cdot & \cos\frac{\widehat{COD}}{2}\ \end{array}\right|\ (\div)\ \Longrightarrow\ \frac{OE}{OF}=\frac{OA\cdot OB}{OA+OB}\ \cdot\ \frac{OC+OD}{OC\cdot OD} \) , intrucat \( \angle AOB\equiv\angle COD \) .
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Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
Corect, insa dupa ce demonstrezi in lucrare la nivel de clasa a saptea relatia folosita (vezi aici).
incat \( O\in (EF) \) . Notam \( x=m(\angle AOE) \) , \( y=m(\angle BOE) \) . Sa se arate ca \( \frac {OE}{OF}=\frac {OA\cdot OB}{OC\cdot OD}\cdot\frac {OD\cdot\sin x+OC\cdot \sin y}{OA\cdot\sin x+OB\cdot\sin y} \) .
Metoda 1 (M.C.). Notam \( \phi =m(\angle AOB) \) . Folosim relatia metrica \( l_a=\frac {2bc\cdot\cos\frac A2}{b+c} \) a lungimii bisectoarei \( l_a \)
intr-un triunghi \( ABC\ : \ \left\|\ \begin{array}{c}
\triangle AOB\ : \ OE=\frac {2\cdot OA\cdot OB\cdot\cos \phi}{OA+OB}\\\\\\\\
\triangle COD\ : \ OF=\frac {2\cdot OC\cdot OD\cdot\cos\phi}{OC+OD}\end{array}\ \right\|\Longrightarrow\ \frac {OE}{OF} \) \( =\frac {OA\cdot OB}{OA+OB}\cdot\frac {OC+OD}{OC\cdot OD} \) .
Metoda 2. Notam \( \left\|\begin{array}{ccc}
M\in BD\ ,\ AM\parallel EF\\\\\\\\
N\in AC\ ,\ DN\parallel EF\end{array}\right\| \) . Se observa ca \( OM=OA \) si \( ON=OD \) . Asadar :
\( \odot\ AM\parallel DN\ \Longrightarrow\ \frac {AM}{DN}=\frac {OA}{ON}=\frac {OA}{OD} \) , adica \( \overline{\underline{\left\|\ \frac {AM}{DN}=\frac {OA}{OD}\ \right\|}}\ (1) \) .
\( \odot\ OE\parallel AM\Longrightarrow\ \frac {OE}{AM}=\frac {OB}{BM}=\frac {OB}{OB+OM}=\frac {OB}{OB+OA}\Longrightarrow\ \overline{\underline{\left\|\ OE=\frac {OB}{OA+OB}\cdot AM\ \right\|}}\ (2) \) .
\( \odot\ OF\parallel DN\Longrightarrow\ \frac {OF}{DN}=\frac {OC}{CN}=\frac {OC}{OC+ON}=\frac {OC}{OC+OD}\Longrightarrow\ \overline{\underline{\left\| OF=\frac {OC}{OC+OD}\cdot DN\ \right\|}}\ (3) \) .
Impartind relatia \( (2) \) la relatia \( (3) \) obtinem relatia \( \frac {OE}{OF}=\frac {OB}{OA+OB}\cdot \frac {OC+OD}{OC}\cdot\frac {AM}{DN} \) .
Tinand seama de relatia \( (1) \) rezulta \( \frac {OE}{OF}=\frac {OB}{OA+OB}\cdot \frac {OC+OD}{OC}\cdot\frac {OA}{OD} \) , adica relatia din concluzie.
Generalizare. Fie patrulaterul convex \( ABCD \) pentru care \( O\in AC\cap BD \) . Consideram \( E\in (AB) \) si \( F\in (CD) \) astfelMetoda 3. Notam \( x=m(\angle AOE)=m(\angle BOE)=m(\angle COF)=m(\angle DOF) \) . Asadar :
\( \odot\ [AOB]=[AOE]+[BOE]\ \Longrightarrow\ OA\cdot OB\cdot\sin 2x=OE\cdot (OA+OB)\cdot\sin x \).
\( \odot\ [COD]=[COF]+[DOF]\ \Longrightarrow\ OC\cdot OD\cdot\sin 2x=OF\cdot (OC+OD)\cdot\sin x \) .
Prin impartirea celor doua relatii se obtine usor relatia din concluzie.
incat \( O\in (EF) \) . Notam \( x=m(\angle AOE) \) , \( y=m(\angle BOE) \) . Sa se arate ca \( \frac {OE}{OF}=\frac {OA\cdot OB}{OC\cdot OD}\cdot\frac {OD\cdot\sin x+OC\cdot \sin y}{OA\cdot\sin x+OB\cdot\sin y} \) .