Răspuns :

Andrutaandruid
a) ABD

D=90,A=30 |=>(T unghi 30) BD=AB/2=x/2

b)ABD

D=90 |=>(TP) x^2=(x/2)^2+AD^2

AD=x radical 3/2

ADC

D=90, C=30|=>(T unghi 30) AD=AC/2

AC=x radical 3

BD/DC=x/2/3x/2=1/3

ADC

D=90|=>(TP)(după unele calcule)DC=3x/2
Targoviste44id

[tex]\it Din \ \ Th.\ \angle\ 30^o \ pentru\ \Delta ABC\ \Rightarrow BC=2AB=2x\\ \\ \widehat{DAB}=\widehat C=30^o\ (au\ acela\c{s}i\ complement, \widehat B).\\ \\ Din \ \ Th.\ \angle\ 30^o \ pentru\ \Delta DAB\ \Rightarrow BD=\dfrac{AB}{2}=\dfrac{x}{2}=0,5x[/tex]

[tex]\it DC=BC-BD=2x-\dfrac{x}{2}=1,5x[/tex]

[tex]\it \dfrac{BD}{DC}=\dfrac{0,5x}{1,5x}=\dfrac{^{10)}0,5}{\ 1,5}=\dfrac{\ 5^{(5}}{15}=\dfrac{1}{3}[/tex]

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