[tex]\it a)\ tg{\widehat{ABD}}=\dfrac{cateta\ \ opus\breve a}{cateta\ \ al\breve a turat\breve a}=\dfrac{AD}{BD}=\dfrac{4}{2}=2[/tex]
[tex]\it b)\ \Delta ABC-dr,\ \hat D=90^0 \stackrel{T.P.}{\Longrightarrow}\ AB^2=AD^2+BD^2=4^2+2^2=16+4=20\\ \\ AB=\sqrt{20}=\sqrt{4\cdot5}=2\sqrt5\ cm[/tex]
c)
[tex]\it Cu\ teorema\ catetei,\ avem:\\ \\ AB^2=BD\cdot BC \Rightarrow BC=\dfrac{AB^2}{BD}=\dfrac{20}{2}=10\ cm[/tex]
