13.
ABC - isoscel, => AB = AC = 6 cm
_____________________________
AB = 6cm
AE = 2cm
_____________________________
AC = 6cm
FC = 4cm
(scazi aceste relatii, si vei obtine:)
=> AF = 2cm
(sau poti scrie ca AC = FC + AF; 6 = 4 + AF; AF = 4cm)
_______________________________
[tex]\frac{AE}{AB} = \frac{AF}{AC} = \frac{2}{6} = \frac{1}{3}[/tex]
Rezulta din reciproca teoremei lui Thales ca,
EF || BC
Cum BC ⊂ α
=> EF || α
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12.
AB = 6cm
AE = 2cm
__________
AC = 18cm
AF = 6cm
__________
[tex]\frac{AE}{AB} = \frac{AF}{AC} = \frac{1}{3}[/tex]
Rezulta din recoproca teoremei lui Thales ca,
Cum BC ⊂ α
=> EF || α
