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-DE}{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
