As shown in the figure, e and F are two points on the diagonal BD of the quadrilateral ABCD, AE ‖ CF, AE = CF, be = DF

As shown in the figure, e and F are two points on the diagonal BD of the quadrilateral ABCD, AE ‖ CF, AE = CF, be = DF

It is proved that ∵ AE ∥ CF ∥ AED = ∠ CFB (3 points) ∵ DF = be, ∵ DF + EF = be + EF, & nbsp; that is, de = BF In △ ade and △ CBF, AE = CF ∠ AED = cfbde = BF (9 points) ≌ △ ade ≌ △ CBF (SAS) ≌ (10 points)