[tex]\it AB\perp BC \Rightarrow\ d(A,\ BC)=AB=6\ cm)\\ \\ \hat A=60^o\ (complementul\ lui\ 30^o\ din\ \Delta BCA )\\ \\ \\ \widehat{ABD}=30^o(complementul\ lui\ 60^o\ din\ \Delta DAB )\\ \\ \\ Th.\ \angle30^o\ pentru\ \Delta BCA \Rightarrow AC=2\cdot AB=2\cdot6=12\ cm\\ \\ Th.\ \angle30^o\ pentru\ \Delta DAB \Rightarrow BD=AB:2=6:2=3\ cm[/tex]
