Resposta:
[tex]\overline{AB}=40\\\\\overline{AC}=80[/tex]
Explicação passo a passo:
Note que pelo caso [tex]AA[/tex] os triângulos [tex]\triangle ADE[/tex] e [tex]\triangle ABC[/tex] são semelhantes.
Disso:
[tex]\frac{\overline{AD}}{\overline{AB}}=\frac{\overline{AE}}{\overline{AC}}[/tex]
Substituindo, teremos:
[tex]\frac{25}{25+(x-5)}=\frac{x+30}{(x+30)+(x+10)}\\\\\frac{25}{25+x-5}=\frac{x+30}{x+30+x+10}\\\\\frac{25}{x+20}=\frac{x+30}{2x+40}\\\\(x+20)\cdot(x+30)=25\cdot(2x+40)\\\\x^2+30x+20x+600=50x+1000\\\\x^2+50x+600=50x+1000\\\\x^2+600=1000\\\\x^2=400\\\\x=\pm20[/tex]
Como [tex]x > 5[/tex], logo [tex]x=20[/tex].
Portanto:
[tex]\overline{AB}=25+20-5=40\\\\\overline{AC}=20+30+20+10=80[/tex]
