解:将△ABD沿AB为对称轴翻折成为△ABE,△ACD沿AC为对称轴翻折成为△ACG,连EB,GC并延长交于F,得△ABD≌△ABE,同理△ACD≌△ACG所以AE=AD=AG,BE=BD=2,CD=CG=3,得正方形AEFG设AE=x,则EF=x,BF=x-2,FC=x-3,在直角三角形BCF中,由勾股定理,得,BC^2=BF^2+CF^2,5^2=(x-2)^2+(x-3)^2解得x=6所以S△ABC=(1/2)*BC*AD=(1/2)*5*6=15