Questões Perdidas ⇒ Solucionário:Racso - Cap XII - Problemas de Geometria y Como Resolverlos - I Edição - Ex:01 Tópico resolvido

[petras]

Problema Proposto
1 - No triângulo ABC se traça a mediana AM e
as cevianas BP e BQ que trissecam o lado AC.
Calcular TK, se AM = 10
[tex3](T: AM \cap BP ~e~ K: AM \cap BQ)[/tex3]

Resposta

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E) 3 (Resposta errada do livro D) 5)

[petras]

[tex3]AD = DE = CE\\
CM = BM\\T. Menelao: \triangle CAM-DB:\\
\underbrace{CD}_{2AD} \cdot AT \cdot BM = AD \cdot MT.\cdot \underbrace{CB}_{2BM}\implies \boxed{AT = MT}\\
\therefore AT = MT = \frac{10}{2}= 5\\
T.Menelao: \triangle CAM-EB:\\
CE \cdot AK \cdot BM = \underbrace{AE}_{2CE} \cdot MK \cdot \underbrace{BC}_{2BM}\implies \boxed{AK = 4MK}\\
AM = AK+MK = 10\implies 4MK+MK = 10 \therefore \boxed{KM=2}\\
KT +KM = 5 \rightarrow 2+ KM = 5 \therefore \boxed{\color{red} KM=3} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]AD = DE = CE\\
CM = BM\\T. Menelao: \triangle CAM-DB:\\
\underbrace{CD}_{2AD} \cdot AT \cdot BM = AD \cdot MT.\cdot \underbrace{CB}_{2BM}\implies \boxed{AT = MT}\\
\therefore AT = MT = \frac{10}{2}= 5\\
T.Menelao: \triangle CAM-EB:\\
CE \cdot AK \cdot BM = \underbrace{AE}_{2CE} \cdot MK \cdot \underbrace{BC}_{2BM}\implies \boxed{AK = 4MK}\\
AM = AK+MK = 10\implies 4MK+MK = 10 \therefore \boxed{KM=2}\\
KT +KM = 5 \rightarrow 2+ KM = 5 \therefore \boxed{\color{red} KM=3} [/tex3]