mateforum.ro

A, B, C, D sunt patru puncte necoplanare astfel \( BA\perp AC \), \( AC\perp DA \), \( DA\perp BC \). Fie \( AG\perp BD \) cu \( G\in BD \) si \( E\in [BC] \) cu proprietatea ca \( AE^2=BE\cdot EC \), iar \( AF\perp DE \) cu \( F\in DE \). Stabiliti daca punctele \( C, F , G \) sunt coliniare.

Stefan Smarandoiu, Rm. Valcea

\( AC\perp DA\Rightarrow DA\perp AC, DA\perp BC \Rightarrow DA\perp(ABC), AB\subset (ABC) \Rightarrow DA\perp AB \).
\( \bigtriangleup {ABC} \) dreptunghic in \( A (BA\perp AC): AE^2=BE\cdot EC\Rightarrow [AE] \) inaltime \( \Rightarrow AE\perp BC \)(Reciproca teoremei inaltimii).
\( \bigtriangleup {BDE}: G\in (BD), F\in (DE), C\in BE \Rightarrow C, F, G \) coliniare \( \Leftrightarrow \frac{CB}{CE}\cdot \frac{FE}{FD}\cdot \frac{GD}{BG}=1 \)(Teorema lui Menelaus).
\( \bigtriangleup{ABC} \) dreptunghic: Aplicand teorema catetei, obtinem \( CE=\frac{AC^2}{BC} \). Atunci \( \frac{CB}{CE}=\frac{CB}{\frac{AC^2}{BC}}=\frac{BC^2}{AC^2} \). Din asemanrea triunghiurilor \( \bigtriangleup{ABC} \) si \( \bigtriangleup{EBA} \), obtinem: \( \frac{BC}{AC}=\frac{AB}{AE} \). Atunci, \( \frac{CB}{CE}=\frac{AB^2}{AE^2} \).
\( \bigtriangleup{ADE} \) dreptunghic: Aplicand teorema catetei, obtinem: \( \frac{FE}{FD}=\frac{\frac{AE^2}{DE}}{\frac{AD^2}{DE}}=\frac{AE^2}{AD^2} \).
\( \bigtriangleup{ABD} \) dreptunghic: Aplicand teorema catetei, obtinem: \( \frac{GD}{BG}=\frac{\frac{AD^2}{BD}}{\frac{AB^2}{DB}}=\frac{AD^2}{AB^2} \).
Atunci \( \frac{CB}{CE}\cdot \frac{FE}{FD}\cdot \frac{GD}{BG}=\frac{AB^2}{AE^2}\cdot\frac{AE^2}{AD^2}\cdot\frac{AD^2}{AB^2}=1\Rightarrow \) punctele \( C, F, G \) sunt coliniare.