答：BC=10,∠D=120°已知觉梯形ABCD中,AD//BC,AB=DC=AD=5,CA⊥AB,可知梯形ABCD为等腰梯形.过A作AE⊥BC交BC于E点,设AE=H,BE=X,则BC=5+2X,CE=5+X,在RT△ABC中,根据勾股定理,得BC^2=AC^2+AB^2（5+2X）^2=AC^2+5^2=AC^2+...