根据勾股定理,在△BCD中,
BC^2+DC^2=BD^2
BC^2=BD^2-DC^2
在△ABC中,
BC^2+AC^2=AB^2
AC^2=AB^2-BC^2=AB^2-(BD^2-DC^2)=(10/3)^2-(2.5^2-1.5^2)=100/9-4
AC≈2.6
根据勾股定理,在△BCD中,
BC^2+DC^2=BD^2
BC^2=BD^2-DC^2
在△ABC中,
BC^2+AC^2=AB^2
AC^2=AB^2-BC^2=AB^2-(BD^2-DC^2)=(10/3)^2-(2.5^2-1.5^2)=100/9-4
AC≈2.6