[tex]\it Fie\ ABC-dr,\ \hjat A=90^o,\ BC=18cm\\ \\ AD-median\breve a\ (D\in BC) \Rightarrow AD=BC:2=18:2=9cm\\ \\\dfrac{\widehat{DAB}}{\widehat{DAC}}=\dfrac{1}{2} \Rightarrow \widehat{DAB}=30^o,\ \ \widehat{DAC}=60^o[/tex]
[tex]\it DA=DB=9\ cm \Rightarrow \Delta DAB-isoscel \Rightarrow \hat B=\widehat{DAB}=30^o\\ \\ Th.\ \angle 30^o\ pentru\ \Delta ABC\Rightarrow AC=BC:2=9\ cm\\ \\ \Delta ABC-dr,\ \hat A=90^o\ \stackrel{T.P.}{\Longrightarrow}\ AB^2=BC^2-AC^2=18^2-9^2=\\ \\ =(18-9)(18+9)=9\cdot27=9\cdot9\cdot3=81\cdot3 \Rightarrow AB=\sqrt{81\cdot3}=9\sqrt3\ cm[/tex]
[tex]\it \mathcal{P}=BC+CA+AB=18+9+9\sqrt3=27+9\sqrt3\ cm[/tex]
