Question

dados um triangulo ABC e um segmento DE com D em AB e E em AC, prove que, se AD:DB= AE:EC, entao DE é paralelo a BC.

Answer 1

Fiz assim:

Se DE for paralelo a BC, com D pertencendo a AB e E pertencendo a ACdo triângulo ABC, o triângulo DEA é semelhante ao ABC

Então:

$\dfrac{AB}{AD}=\dfrac{AC}{AE}$

Olhando pra figura, podemos ver que AB = AD + DB e AC = AE + EC:

$\dfrac{AB}{AD}=\dfrac{AC}{AE}~~\therefore~~\dfrac{AD+DB}{AD}=\dfrac{AE+EC}{AE}\\\\\\\dfrac{AD+DB}{AD}=\dfrac{AE+EC}{AE}~~\therefore~~\dfrac{AD}{AD}+\dfrac{BD}{AD}=\dfrac{AE}{AE}+\dfrac{EC}{AE}\\\\\\\dfrac{AD}{AD}+\dfrac{BD}{AD}=\dfrac{AE}{AE}+\dfrac{EC}{AE}~~\therefore~~1+\dfrac{BD}{AD}=1+\dfrac{EC}{AE}\\\\\\1+\dfrac{BD}{AD}=1+\dfrac{EC}{AE}~~\therefore~~\dfrac{BD}{AD}=\dfrac{EC}{AE}$

$\dfrac{BD}{AD}=\dfrac{EC}{AE}~~\therefore~~\boxed{\boxed{\dfrac{AD}{BD}=\dfrac{AE}{EC}}}$

Answer 2

Como é paralelo podemos resolver usando semelhança dos triângulos.

Sendo:

$\frac{AB}{AD} = \frac{AC}{AE } \\ \\ \frac{AD + BD}{AD} = \frac{AE + CE}{AE} \\ \\ 1 + \frac{BD}{AD} = 1 + \frac{AE + CE}{AE} \\ \\ \frac{BD}{AD} = \frac{CE}{AE} \\ \\ \\ \frac{AD}{BD} = \frac{AE}{CE}$