2. Take two right-angled sides AB and BC of right-angled triangle ABC as one side, and make equilateral triangle ABE and equilateral triangle BCF connect EF and EC outward respectively. Explain (1)EF=EC(2)EB⊥CF?

1. Solution: P is the vertical segment PE and PF of AB and CD respectively.

∫S△PAB = S△PCD (known)

∴ 1/2*ab*pe= 1/2*cd*pf

AB = CD (known)

∴PE=PF (Properties of Equation)

∴OP bisector ∠AOD (points with equal distance to both sides of the corner are on the bisector of the corner)

2. Solution: (1) ∵∠ CBE = ∠ CBA+∠ Abe = 90+60 =150,

∠FBE = 360-∠FBC-∠CBA-∠ Abbe =360 -60 -90 -60 = 150

∴∠CBE=∠FBE (equivalent substitution)

At △CBE and △FBE,

Common edge

BF=BC (three sides of an equilateral triangle are equal)

∠CBE=∠FBE (certification)

∴△CBE≌△FBE(SAS)

∴EF=EC (corresponding sides are equal)

(2) √△ CBE△ FBE (certification)

∴∠BEF=∠BEC (corresponding angles are equal),

EF=EC (corresponding sides are equal)

Extend EB and cross FC at G point.

At △CEG and △FEG,

GE=GE (male side)

∠BEF=∠BEC (certification)

EF=EC (certification)

∴△CEG≌△FEG(SAS)

∴∠CGE=∠FGE (corresponding angles are equal)

∠∠CGE+∠FGE = 180 (definition of adjacent complementary angle)

∴ 2 ∠ CGE = 180 (equivalent substitution)

∴∠ CGE = 90 (the nature of the equation)

∴EB⊥CF (vertical definition)