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The fourth problem of mathematics in grade three
Solution: the intersection D is DG⊥DE, and the BC extension line is at G.

∵ quadrilateral ABCD is a square,

∴AD=CD,∠DAE=∠ADC=∠DCG=90 .

∴∠ADE=∠CDG (complementary angles of the same angle are equal).

∴rtδdae≌rtδcdg(asa)。

∴AE=CG,DE=DG。

∫∠EDF = 45,

∴∠ADE+∠CDF=∠CDG+∠CDF=∠GDF=45 .

DF = DF,

∴δdef≌δdgf(sas)。

∴EF=FG=6。

∫FG = CG+CF = AE+CF,

∴EF=AE+CF=6。