As shown in the figure, the quadrilateral ABCD is a diamond, f is the intersection of a point DF on AB, AC at e. the angle AFD = angle CBE is proved  

As shown in the figure, the quadrilateral ABCD is a diamond, f is the intersection of a point DF on AB, AC at e. the angle AFD = angle CBE is proved  

∫ quadrilateral ABCD is rhombic ∫ ACB = ∠ ACD, BC = CD, EC = EC ≌ BCE ≌ DCE ≌ CDE = ∠ CBE ∫ ab ∥ CD ∫ AFD = ∠ CDE ∫ CDE = ∠ CBE