Problema cu bisectoare
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BogdanCNFB
- Thales
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Problema cu bisectoare
In \( \triangle{ABC} \), bisectoarea \( \angle A \) intersecteaza latura \( (BC) \) in \( D \). Demonstrati ca \( AD<\sqrt{AB\cdot AC} \).
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Virgil Nicula
- Euler
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Notam cel de-al doilea punct \( S \) unde bisectoarea \( [AD \) taie cercul circumscris \( w \) al triunghiului dat. Se observa ca
\( \triangle ABS\sim\triangle ADC \) , adica \( \frac {c}{AD}=\frac {AS}{b} \) de unde obtinem \( AS\cdot AD=bc \) , adica \( AD\cdot (AD+DS)=bc \) . Asadar,
folosind puterea punctului \( D \) in raport cu cercul \( w \) , adica \( DA\cdot DS=DB\cdot DC \) , obtinem \( \overline{\underline {\left\|\ AD^2=bc-DB\cdot DC\ \right\|}} \) .
\( \triangle ABS\sim\triangle ADC \) , adica \( \frac {c}{AD}=\frac {AS}{b} \) de unde obtinem \( AS\cdot AD=bc \) , adica \( AD\cdot (AD+DS)=bc \) . Asadar,
folosind puterea punctului \( D \) in raport cu cercul \( w \) , adica \( DA\cdot DS=DB\cdot DC \) , obtinem \( \overline{\underline {\left\|\ AD^2=bc-DB\cdot DC\ \right\|}} \) .
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mihai miculita
- Pitagora
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- Location: Oradea, Romania
\( \mbox{Intrucat: } \sqrt{|BD|.|DC|}\le\frac{|BD|+|DC|}{2}=\frac{|BC|}{2}=\frac{a}{2}\Rightarrow |BD|.|DC|\le \frac{a^2}{4}\Rightarrow |AD|^2\ge bc-\frac{a^2}{4};\\
\mbox{ asa ca putem complecta inegalitatea cu: } \frac{\sqrt{4bc-a^2}}{2}\le |AD|.\\
\mbox{Evident, egalitatea are loc doar in cazul in care |BD|=|DC|\Leftrightarrow |AB|=|AC|(\mbox{in cazul de fata}). \)
\mbox{ asa ca putem complecta inegalitatea cu: } \frac{\sqrt{4bc-a^2}}{2}\le |AD|.\\
\mbox{Evident, egalitatea are loc doar in cazul in care |BD|=|DC|\Leftrightarrow |AB|=|AC|(\mbox{in cazul de fata}). \)