Re: Problema 41 - Proporcionalidade e Semelhança -Vol. 7
Enviado: 31 Mai 2025, 17:29
[tex3]AM = x: MD = y\\
\triangle MOD \cong \triangle NOD \implies NO = MO = \frac{MN}{2} = 1
\\\triangle ADB:
(x+y)^2+AB^2 = DB^2\\
(x+y)^2 + 3=DB^2(i)\\
\triangle MDO \sim \triangle BDA:
\frac{MO}{AB} = \frac{\frac{DB}{2}}{x+y} \\
\frac{1}{\sqrt3} = \frac{\frac{DB}{2}}{x+y} \implies x+y = \frac{\sqrt3DB}{2}\\
De(i):\frac{3DB^2}{4}+3 = DB^2 \implies DB^2 = 12\\
\therefore (x+y)^2+3 = 12 \implies \boxed{x+y = AD = 3}
[/tex3]
\triangle MOD \cong \triangle NOD \implies NO = MO = \frac{MN}{2} = 1
\\\triangle ADB:
(x+y)^2+AB^2 = DB^2\\
(x+y)^2 + 3=DB^2(i)\\
\triangle MDO \sim \triangle BDA:
\frac{MO}{AB} = \frac{\frac{DB}{2}}{x+y} \\
\frac{1}{\sqrt3} = \frac{\frac{DB}{2}}{x+y} \implies x+y = \frac{\sqrt3DB}{2}\\
De(i):\frac{3DB^2}{4}+3 = DB^2 \implies DB^2 = 12\\
\therefore (x+y)^2+3 = 12 \implies \boxed{x+y = AD = 3}
[/tex3]