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Fie ABC un triunghi oarecare, E apartine (AB), F apartine (AC). Aratati ca EF paralel cu BC daca:
a) AE=12 ; EB=18 ; AF=10; FC=15
b) AE=4 ; AB=16 ; AC=20 ; FC=15
c) AB=28; EB=21; AF=6 ; FC=18
SE FACE PRIN METODA LU THALES! DAU 30 PUNCTE!


Răspuns :

a)AE/EB=AF/AC <=> 12/18=10/15 <=>2/3=2/3 (Adevarat)  
b)
AC=20
FC=15=>AF=5   AE/AB=AF/AC <=>4/16 =5/20 <=>1/4=1/4 (A)

C)AB=28
EB=21=>AE=28-21=7 
AE/EB=AF/FC <=>7/21=6/18 <=>1/3=1/3 (A)

la toate aplici reciproca lui Thales...

a) AB/AE = AC/AF => 30/12 = 25/10 => 30*10 = 12*25 => 300 = 300

b) AB/AE = AC/AF => 16/4 = 20/5 => 16*5 = 4*20 => 80 = 80

c) AB/AE = AC/AF => 28/7 = 24/6 => 7*24 = 6*28 => 168 = 168