Aplicatie la T.Ceva si T.Menelaus ex.1
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Aplicatie la T.Ceva si T.Menelaus ex.1
Se considera triunghiul \( ABC \) neisoscel si \( [AD, [BE, [CF \) bisectoarele exterioare ale unghiurilor triunghiului, \( D \in BC, E \in AC, F \in AB \). Sa se demonstreze ca punctele \( D, E, F \) sunt coliniare.
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Omer Cerrahoglu
- Euclid
- Posts: 34
- Joined: Mon Mar 17, 2008 1:08 pm
Presupunem ca \( D \in (CB \), \( E \in (CA \) si \( F \in (BA \).
Avem binecunoscuta proprietate ca \( \frac{DB}{DC}=\frac{A[ABD]}{A[ACD]}(1) \)
Avem ca \( A[ABD]=\frac{AD.AB.\sin{DAB}}{2}=\frac{AD.AB.\sin{\frac{B+C}{2}}}{2} (2) \)
Analog \( A[ACD]=\frac{AD.AC.\sin{(A+\frac{B+C}{2})}}{2}=\frac{AD.AC.\sin{(180^0-A-\frac{B+C}{2})}}{2}=\frac{AD.AC.\sin{\frac{B+C}{2}}}{2} (3) \)
Din (1), (2) si (3) avem ca \( \frac{DB}{DC}=\frac{AB}{AC}(*) \)
Analog \( \frac{CE}{EA}=\frac{BC}{AB}(**) \) si \( \frac{AF}{BF}=\frac{AC}{BC}(***) \)
Din (*), (**), (***) avem ca \( \frac{DB}{DC}.\frac{CE}{EA}.\frac{AF}{BF}=\frac{AB}{AC}.\frac{BC}{AB}.\frac{AC}{BC}=1 \)
Deci, din reciproca teoremei lui Menelaus avem ca punctele \( D, E, F \) sunt colineare.
Avem binecunoscuta proprietate ca \( \frac{DB}{DC}=\frac{A[ABD]}{A[ACD]}(1) \)
Avem ca \( A[ABD]=\frac{AD.AB.\sin{DAB}}{2}=\frac{AD.AB.\sin{\frac{B+C}{2}}}{2} (2) \)
Analog \( A[ACD]=\frac{AD.AC.\sin{(A+\frac{B+C}{2})}}{2}=\frac{AD.AC.\sin{(180^0-A-\frac{B+C}{2})}}{2}=\frac{AD.AC.\sin{\frac{B+C}{2}}}{2} (3) \)
Din (1), (2) si (3) avem ca \( \frac{DB}{DC}=\frac{AB}{AC}(*) \)
Analog \( \frac{CE}{EA}=\frac{BC}{AB}(**) \) si \( \frac{AF}{BF}=\frac{AC}{BC}(***) \)
Din (*), (**), (***) avem ca \( \frac{DB}{DC}.\frac{CE}{EA}.\frac{AF}{BF}=\frac{AB}{AC}.\frac{BC}{AB}.\frac{AC}{BC}=1 \)
Deci, din reciproca teoremei lui Menelaus avem ca punctele \( D, E, F \) sunt colineare.