Questões Perdidas ⇒ Solucionário:Racso - Cap XVI - Problemas de Geometria y Como Resolverlos - I Edição - Ex:25 Tópico resolvido

[petras]

Problema Proposto
25 - No triángulo ABC: [tex3]\angle A = 2\angle C[/tex3], IG|| AC,
sendo I o incentro e G o baricentro. Se IG =a, calcular
a distância ID se BD é bissetriz interior.

Resposta

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D) 3a[tex3]\sqrt{2}[/tex3]

[petras]

Sendo IG = x
[tex3]\mathsf{ IG∥AC⟹(propriedade): \boxed{2b=a+c}(I)\\

△AIB∼△CDB \implies \frac{BI}{c}=\frac{BD}{a}\\

BI=\frac{2BD}{3} em(I) \implies\\
c=\frac{2a}{3}
\implies a=\frac{6b}{5} ~e ~c=\frac{4b}{5}\\

\frac{AD}{c}=\frac{CD}{a}\\
\triangle BIG \sim \triangle BDM \implies\\
\frac{BI}{BD} = \frac{x}{DM}\rightarrow \frac{2}{3} = \frac{x}{DM} \therefore DM = \frac{3x}{2}\\
\therefore \frac{a}{c}=\frac{3}{2}=\frac{CD}{AD}=\frac{\frac{b}{2}+\frac{3x}{2}}{\frac{b}{2}-\frac{3x}{2}}=\frac{b+3x}{b−3x}⟹b=15x\\

Fórmula~Bissetriz:\\

BD^2={ac(1−\frac{b^2}{(a+c)^2})=\frac{3ac}{4}=162x^2⟹ BD=9x\sqrt2 \\
ID=\boxed{\color{red}\frac{BD}{3}=3x\sqrt2}}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{ IG∥AC⟹(propriedade): \boxed{2b=a+c}(I)\\

△AIB∼△CDB \implies \frac{BI}{c}=\frac{BD}{a}\\

BI=\frac{2BD}{3} em(I) \implies\\
c=\frac{2a}{3}
\implies a=\frac{6b}{5} ~e ~c=\frac{4b}{5}\\

\frac{AD}{c}=\frac{CD}{a}\\
\triangle BIG \sim \triangle BDM \implies\\
\frac{BI}{BD} = \frac{x}{DM}\rightarrow \frac{2}{3} = \frac{x}{DM} \therefore DM = \frac{3x}{2}\\
\therefore \frac{a}{c}=\frac{3}{2}=\frac{CD}{AD}=\frac{\frac{b}{2}+\frac{3x}{2}}{\frac{b}{2}-\frac{3x}{2}}=\frac{b+3x}{b−3x}⟹b=15x\\

Fórmula~Bissetriz:\\

BD^2={ac(1−\frac{b^2}{(a+c)^2})=\frac{3ac}{4}=162x^2⟹ BD=9x\sqrt2 \\
ID=\boxed{\color{red}\frac{BD}{3}=3x\sqrt2}}}[/tex3]

(Solução:MathLover)