As shown in the figure, it is known that the quadrilateral ABCD is a diamond, f is the intersection of a point DF on AB, AC is on e, and it is proved that ∠ AFD = ∠ CBF

As shown in the figure, it is known that the quadrilateral ABCD is a diamond, f is the intersection of a point DF on AB, AC is on e, and it is proved that ∠ AFD = ∠ CBF

It is proved that: ∵ - CBE is the outer angle of △ BFE (known) ∵ - CBE = ∵ - bef + ∵ - BFE (an outer angle of triangle is equal to the sum of two non adjacent inner angles). ∵ - AFD = ∵ - BFE (known) ∵ - AFD = ∵ CBE (equivalent substitution) it is proved that connecting BD, because the quadrilateral ABCD is a diamond, AC vertically bisects BD, BD bisects ∵ ABC, x0d so be = De, so ∵ - EBD = ∵ EDB, ∵ CBD =