Se considera triunghiul ABC si (AD bisectoarea unghi BAC,D ∈BC si DE paralel AB,E∈AC.Daca AB 12 cm si AC 16 cm,calculati AE,DE si CE
rezolvare prin teorema lui thales
va rog!

Daca ED||AB, atunci <EDA≡<DAB (alterne interne) 
Dar , din Ipoteza stim ca <EAD≡<DAB=> 
<EAD≡<EDA⇒ΔEAD=isoscel ⇒ AE≡ED
Conform Thales:
[tex] \frac{CE}{CA} = \frac{DE}{AB} [/tex]
[tex] \frac{CA-AE}{CA} = \frac{DE}{AB} [/tex]
Dar AE=DE⇒
[tex] \frac{CA-DE}{CA} = \frac{DE}{AB} [/tex]
[tex] \frac{16-AE}{16} = \frac{DE}{12} [/tex]
12(16-DE)=16DE
192-12DE=16DE
192=28DE
DE=192/28
DE=48/7=AE
CE=AC-AE=16-48/7=112/7-48/7=64/7
CE=64/7