Bisectoare si mediane
Moderators: Bogdan Posa, Laurian Filip
Bisectoare si mediane
In triunghiul oarecare ABC, bisectoarele (BE si (CF taie mediana AD in punctele M si N. Sa se arate ca \( \frac{AM}{MD}+\frac{AN}{ND}>2 \).
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Andi Brojbeanu
- Bernoulli
- Posts: 294
- Joined: Sun Mar 22, 2009 6:31 pm
- Location: Targoviste (Dambovita)
In triunghiul \( ABD \), din teorema bisectoarei rezulta \( \frac{AM}{MD}=\frac{AB}{BD} \).
In triunghiul \( ADC \), din teorema bisectoarei rezulta \( \frac{AN}{ND}=\frac{AC}{DC} \).
Cum AD mediana \( \Rightarrow BD=DC \).
Atunci \( \frac{AM}{MD}+\frac{AN}{ND}=\frac{AB}{BD}+\frac{AC}{DC}=\frac{AB+AC}{2BD}=\frac{AB+AC}{BD} \).
Dar \( \frac{AB+AC}{BD}>2\Leftrightarrow AB+AC>2BD, AB+AC>BC \), adevarat.
In triunghiul \( ADC \), din teorema bisectoarei rezulta \( \frac{AN}{ND}=\frac{AC}{DC} \).
Cum AD mediana \( \Rightarrow BD=DC \).
Atunci \( \frac{AM}{MD}+\frac{AN}{ND}=\frac{AB}{BD}+\frac{AC}{DC}=\frac{AB+AC}{2BD}=\frac{AB+AC}{BD} \).
Dar \( \frac{AB+AC}{BD}>2\Leftrightarrow AB+AC>2BD, AB+AC>BC \), adevarat.