过A做AE┻BC,E为垂足.
则直角三角形CAE中,PD//AE,且P为AB中点,所以CD=DE=BE3.由勾股定理,AE^2=AC^2-(CD+DE)^2
直角三角形BAE中,AB^2=AE^2+BE^2
即AB^2=AC^2-36+9=AC^2-27 (1)
直角三角形ABC中,AB^2=BC^2-AC^2 (2)
(1)+(2)得,
2AB^2=54
AB^2=27