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20 16 Handan Yimo mathematics
(1) proof: in rhombic ABCD, AB=BC,

∫BG average score ∠ABC,

∴∠ABG=∠CBG,

At △ABG and △CBG,

AB=BC∠ABG=∠CBGBG=BG,

∴△ABG≌△CBG(SAS),

∴ag=cg;

(2) Proof: Connect alternating current,

∫AC is the diagonal of diamond ABCD,

∴∠ACE=∠ACF,

In △ACE and △ACF,

CE=CF∠ACE=∠ACFAC=AC,

∴△ACE≌△ACF(SAS),

∴∠CAE=∠CAF,

According to (1), AG=CG,

∴∠CAE=∠ACG,

∴∠ACG=∠CAF,

∴cg∥af;

(3)△ Abe ∽△BGE.

The reasons are as follows: (1) indicates △ abg △ cbg.

∴∠BAG=∠BCG,

BG = CG,

∴∠CBG=∠BCG,

∴∠BAG=∠CBG,

∫∠AEB =∠ Berg,

∴△ABE∽△BGE.