Mathematical geometry problems in the eighth grade last semester
AD is parallel to BC, BD is perpendicular to AD, so BD is perpendicular to BC, triangle ADB and triangle BDC are right triangles, and E and F are midpoint, so DE= semi-AB, DC= semi-CD, and DE=BF, so AB=∠A=C, which proves that triangle DEB and triangle Bd F are congruent, so DF is parallel to BE, so abcd is parallelogram, so ∞.