BD este bisectoarea <ADC => m(< ADO)=60
BD=2BO => BO=BD/2 => BO=4√2/2 => BO=2√2
Dar BO=DO => DO=2√2
In ΔADO,m(<AOD)=90 =>tgD=AO/OD =>tg60=AO/2√2 =>√3=AO/2√2 => AO=2√6
AO=AC/2 ,deci AC=2AO => AC=4√6
In ΔAOD,cos60=OD/AD => 1/2=2√2/AD => AD=4√2
Perimetrul=4AD=16√2