Se considera triunghiul isoscel \( ABC(AB=AC) \). Punctul \( E \) este situat pe perpendiculara in \( B \) pe \( AB \), iar punctul \( F \) este situat pe perpendiculara in \( C \) pe \( AC \), astfel incat \( E \) si \( F \) sunt in exteriorul unghiului \( BAC \), iar \( BE=CF \). Fie \( \{O\}=BF\cap CE \) si \( \{D\}=BE\cap CF \).
Sa se arate ca \( BF=CE; OB=OC; DB=DC \).
O.VI.31
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Andi Brojbeanu
- Bernoulli
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\( \triangle ABC \) isoscel \( \Longrightarrow \)\( m(\angle ABC)=m(\angle ACB)(1) \)
\( m(\angle EBC)=90+m(\angle ABC)(2) \)
\( m(\angle FCB)=90+m(\angle ACB)(3) \)
-din \( (1),(2),(3)\Longrightarrow \angle EBC\equiv \angle FCB \)
\( \triangle EBC\equiv\triangle FCB\Longrightarrow BF=CE, \angle ECB\equiv\angle FBC \)
\( \angle OCB\equiv \angle OBC(E-O-C,F-O-B)\Longrightarrow \triangle OBC \) isoscel, \( OB=OC \).
\( \angle DBC=180-\angle EBC(4) \)
\( \angle DCB=180-\angle FCB(5) \)
-din \( (3),(4),(5)\Longrightarrow \angle DBC\equiv \angle DCB\Longrightarrow \triangle DBC \) isoscel, \( DB=DC \).
\( m(\angle EBC)=90+m(\angle ABC)(2) \)
\( m(\angle FCB)=90+m(\angle ACB)(3) \)
-din \( (1),(2),(3)\Longrightarrow \angle EBC\equiv \angle FCB \)
\( \triangle EBC\equiv\triangle FCB\Longrightarrow BF=CE, \angle ECB\equiv\angle FBC \)
\( \angle OCB\equiv \angle OBC(E-O-C,F-O-B)\Longrightarrow \triangle OBC \) isoscel, \( OB=OC \).
\( \angle DBC=180-\angle EBC(4) \)
\( \angle DCB=180-\angle FCB(5) \)
-din \( (3),(4),(5)\Longrightarrow \angle DBC\equiv \angle DCB\Longrightarrow \triangle DBC \) isoscel, \( DB=DC \).