[tex]\it m(\widehat{C})=60^o\ (complementul\ lui\ 30^o)\\ \\ Pentru\ \ \Delta ACD\ avem:\ m(\widehat{ADC})= 180^o-(m(\widehat{CAD})+m(\widehat{C}))=\\ \\ =180^o-(60^o+60^o)=60^o \Rightarrow \Delta ACD-echilateral \Rightarrow\\ \\ \Rightarrow \mathcal{P}_{ACD}=3\cdot AC=3\cdot4=12\ cm[/tex]
