a. in triunghiul ADB: m(D)=90
pitagora: AD^2=AB^2-BD^2=81-9=72=>AD=6rad2
triunghiul ABC: AC^2=BC^2-AB^2=(3+CD)^2-81=CD^2+6CD+9-81=CD^2+6CD-72
triunghiul CAD: AC^2=CD^2+AD^2=CD^2+72
egalam ecuatiile si, dupa reduceri avem: 6CD=144=>CD=24
BC=CD+DB=24+3=27
b.AB^2=BC^2-AC^2=256-64=192=>AB=8rad3
AB*AC=AD*BC=>AD=AC*AB/BC=4rad3
in triunghiul ACD: CD^2=AC^2-AD^2=64-48=16=>CD=4
c.in triunghiul ADB: AD^2=AB^2-BD^2=AB^2-441
in triunghiul ACD: AD^2=AC^2-CD^2=BC^2-AB^2-3969=3087-AB^2
egaland ecuatiile obtinem: 2*AB^2=3087+441=3528=>AB^2=1764=>AB=42
AC^2=7056-1764=5292=>AC=42rad3
