a)fie AM_|_CD=> AM=inaltime=2√3
sin ADM=AM/AD
√2/2=2√3/AD
AD=2√6
b)daca <ADM=45=> ΔADM =isoscel si AD=AM=2√3
fie BN_|_CD
In ΔBNC avem
tg BCN=tg 30=√3/3
√3/3=BN/NC
NC=6√3/√3=6
AB+MN=6
CD=DM+MN+NC=2√6+6+6=2√3+12=2(√3+6)
c)in ΔBCN
sin <BCN=sin 30=1/2=BN/BC
1/2=2√3/BC
BC=4√3
in ΔDBC
distanta de la D la latura BC este perpendiculara dusa din D pe BC
fie DP_|_BC
CD*BN=BC*DP
2(√3+6)*2√3=DP*4√3
DP=2(√3+6)*2√3/4√3=√3+6
