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In triunghiul ABC, consideram D€(AB) si construim DE||BC si DF||AC, unde E€(AC), iar F€(BC). Demonstrati ca DF÷AC +DE÷BC=1

Răspuns :

DE ║BC ⇒ [tex] \frac{AD}{AB} [/tex]=[tex] \frac{AE}{AC} [/tex] = [tex] \frac{DE}{BC} [/tex] 
DF ║AC ⇒ [tex] \frac{BD}{AB} [/tex] = [tex] \frac{BF}{BC} [/tex] = [tex] \frac{DF}{AC} [/tex] 
[tex] \frac{DF}{AC} [/tex] + [tex] \frac{DE}{BC} [/tex] = 1 
[tex] \frac{AD}{AB} + \frac{BD}{AB} = 1 [/tex]
[tex] \frac{AD + BD}{AB} = 1 [/tex]
[tex] \frac{AB} { AB } = 1 [/tex]
1 = 1 ( A ) 

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