E:5756: Difference between revisions

Revision as of 19:04, 11 December 2024

E:5756 (Dumitru Acu)

Fie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ABCD} un romb. Prin vârful Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} ducem o dreaptă arbitrară care intersectează pe Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle BC} în Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} , pe Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle DC} în Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F} , iar pe diagonala Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle BD} în Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G} . Să se arate că dreapta Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle CG} este tangentă în Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} cercului circumscris triunghiului Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ECF} .

Soluție

Din faptul că semidreapta Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (DB} este bisectoarea unghiului Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sphericalangle ADC} și semidreapta Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (GB} este bisectoarea unghiului Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sphericalangle AGC} se deduce că Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ AG \right] \equiv \left[ CG \right]}

În triunghiul Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle FGC} , aplicăm Teorema bisectoarei, pentru bisectoarea $(GD$ a unghiului Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sphericalangle FGC} și obținemFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{CD}{DF} = \frac{GC}{GF}} Cum patulaterul Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ABCD} este un romb, avem Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle AB \parallel BC} , deci Teorema lui Thales implică Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{CD}{DF}=\frac{EA}{AF}} Atunci avemFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{EA}{AF} = \frac{GA}{GF} \Rightarrow \frac{EA}{GA} = \frac{AF}{GF}} Prin intermediul proporțiilor derivate se obține

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{GA-AE}{GA} = \frac{GF - FA}{GF} \Rightarrow \frac{GE}{GA} = \frac{GA}{GF}} ceea ce revine laFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle GA^2 = GE \cdot GF \Leftrightarrow GC^2 = GE \cdot GF}

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 Din faptul că semidreapta <math>(DB</math> este bisectoarea unghiului <math>\sphericalangle ADC</math> și semidreapta <math>(GB</math> este bisectoarea unghiului <math>\sphericalangle AGC</math> se deduce că <math display="block" id="1e5756">\left[ AG \right] \equiv \left[ CG \right]</math>
-[[File:Gm 1-1977 e-5756.png|thumb|left]]În triunghiul <math>FGC</math>, aplicăm [https://ro.wikipedia.org/wiki/Teorema_bisectoarei Teorema bisectoarei], pentru bisectoarea <math>(GD</math> a unghiului <math>\sphericalangle FGC</math> și obținem<math display="block" id="2e5756">\frac{CD}{DF} = \frac{GC}{GF}</math>Cum patulaterul <math>ABCD</math> este un romb, avem <math>AB \parallel BC</math>, deci [https://ro.wikipedia.org/wiki/Teorema_lui_Thales Teorema lui Thales] implică <math display="block" id="3e5756">\frac{CD}{DF}=\frac{EA}{AF}</math>Atunci avem
+[[File:Gm 1-1977 e-5756.png|thumb|left]]În triunghiul <math>FGC</math>, aplicăm [https://ro.wikipedia.org/wiki/Teorema_bisectoarei Teorema bisectoarei], pentru bisectoarea <math>(GD</math> a unghiului <math>\sphericalangle FGC</math> și obținem<math display="block" id="2e5756">\frac{CD}{DF} = \frac{GC}{GF}</math>Cum patulaterul <math>ABCD</math> este un romb, avem <math>AB \parallel BC</math>, deci [https://ro.wikipedia.org/wiki/Teorema_lui_Thales Teorema lui Thales] implică <math display="block" id="3e5756">\frac{CD}{DF}=\frac{EA}{AF}</math>Atunci avem<math display="block">\frac{EA}{AF} = \frac{GA}{GF} \Rightarrow \frac{EA}{GA} = \frac{AF}{GF}</math>Prin intermediul proporțiilor derivate se obține
+<math display="block">\frac{GA-AE}{GA} = \frac{GF - FA}{GF} \Rightarrow \frac{GE}{GA} = \frac{GA}{GF}</math>ceea ce revine la<math display="block">GA^2 = GE \cdot GF \Leftrightarrow GC^2 = GE \cdot GF</math>