It is known that: as shown in the figure, AC bisects ∠ bad, CE ⊥ AB in E & nbsp; CF ⊥ ad in F, and BC = DC

It is known that: as shown in the figure, AC bisects ∠ bad, CE ⊥ AB in E & nbsp; CF ⊥ ad in F, and BC = DC

It is proved that ∵ AC bisects ∠ bad, CE ⊥ AB in E & nbsp; CF ⊥ ad in F, ∵ f = ∠ CEB = 90 ° CE = cf. in RT △ CEB and RT △ CFD, BC = DCCE = CF, ≌ CEB ≌ △ CFD (HL), ≌ be = DF

As shown in the figure, in △ ABC, BD = CE, DF = EF, ab = AC

It is proved that: as shown in the figure, if crossing point D as DH ‖ AC intersects BC with H, then ∠ e = ∠ HDF, in △ DFH and △ EFC, ∠ e = ∠ hdfdf = EF ∠ DFH = ∠ EFC, ｜ △ DFH ≌ EFC (ASA), ≌ DH = CE, ≌ BD = CE, ≌ BD = DH, ∥ B = DH, ∥ Bhd, ∥ DH ‖ AC, ≌ Bhd = ∠ ACB, ≌ B = AC

Equilateral △ ABC, a point D on the edge of AB, extend BC to e, intersect AC with F, CE equals ad, and prove that DF = EF

A point D on the edge of equilateral △ ABC, AB, extends BC to e so that CE = ad, de intersects AC with F, proving DF = EF
It is proved that △ ADG is an equilateral triangle if DG ‖ BC intersects AC with G
∴DG=AD=CE
Easy syndrome △ FDG ≌ △ FEC
∴DF=EF

As shown in the figure, ab = CD, AF = CE, be ⊥ AC in E, DF ⊥ AC in F, try to explain △ Abe ≌ △ CDF

Because EF = EF, AE = CF
If the two corresponding sides of a right triangle are equal, the two triangles are congruent