Question

In ΔABC se duce AD bisectoare unghiului A, D∈[BC], DE||AB,E∈[AC].Daca AB=20cm, AC=30cm, BC=25cm, sa se afle perimetrul ΔDEC.

REZOLVAREA COMPLETA!!!! NU DOAR RASPUNSURILE!!

Answer 1

Aplicam teorema bisectoarei:

$\dfrac{DB}{DC}=\dfrac{AB}{AC} \Rightarrow \dfrac{DB}{DC}=\dfrac{20}{30} \Rightarrow \ \dfrac{DB}{DC}=\dfrac{2}{3}$

Derivam ultima proportie si obtinem :

[tex]\dfrac{DB+DC}{DC} =\dfrac{2+3}{3} \Rightarrow \dfrac{BC}{DC}=\dfrac{5}{3} \Rightarrow \dfrac{25}{DC}=\dfrac{5}{3} \Rightarrow DC=\dfrac{25\cdot3}{5}\Rightarrow \\\;\\ \Rightarrow\ DC = 15 cm[/tex]

Dar, ED||AB si  aplicand teorema fundamentala a asemanarii, rezulta:

ΔEDC ~ ΔABC ⇒ raportul de asemanare este:
 
                                  k = DC/BC =15/25=3/5.

Stim ca rapoartele perimetrelor celor doua triunghiuri asemenea
 
este egal cu raportul k de asemanare, adica:

$\dfrac{\mathcal{P}_{EDC}}{\mathcal{P}_{ABC}} =\dfrac{3}{5} \ \ \ \ (1)$

[tex]\mathcal{P}_{ABC} = 15+25+30 =75\ cm\ \ \ \ (2) \\\;\\ Din\ relatiile\ (1),\ (2) \Rightarrow\ \dfrac{\mathcal{P}_{EDC}}{75} =\dfrac{3}{5} \Rightarrow\ \mathcal{P}_{EDC} =\dfrac{75\cdot3}{5} \Rightarrow\ \mathcal{P}_{EDC} = 45 \\\;\\ .[/tex]