已知矩形ABCD.以AD,AB为边向内做等边三角形ADE和等边三角形ABF,延长DF,BE相交于点G(1)求证：DF=BE.(2)猜想∠EGF的度数,并说明理由.（3）当点G位于对角线AC上时,求证∠DGA=∠BGA并直接写出GE与BE的数

Question

已知矩形ABCD.以AD,AB为边向内做等边三角形ADE和等边三角形ABF,延长DF,BE相交于点G
(1)求证：DF=BE.(2)猜想∠EGF的度数,并说明理由.（3）当点G位于对角线AC上时,求证∠DGA=∠BGA并直接写出GE与BE的数量关系.图交代的很清楚,可自行脑部.

Answer 1

（1）证明：∵△ADE和△ABF都是正三角形,∴AF=AB,AD=AE,∠DAE=∠FAB=60°,
∵四边形ABCD是矩形,∴∠DAB=90°,∴∠DAF=∠EAB=30°,∴△ADF≌△ABE(SAS) ∴DF＝BE „„„„„„„„„„„„„„„„„„„„„„„„„„3分 （2）∠EGF=120°,„„„„„„„„„„„„„„„„„„„„„„„„4分 ∠EGF＝∠DFB-∠FBG =60°+∠AFD-∠FBG =60°+∠ABF-∠FBG 60°+60°+∠FBG -∠FBG =120°,„„„„„„„„„„„6分 （3）①过点A作AM⊥DG,AN⊥BG于点M、N,
∵△ADF≌△ABE（已证）,∴∠DFA=∠EBA,AF=AB,又∵∠FMA=∠BNA=90°,∴△AMF≌△ABN,„„„„„„„„„„„„8分 ∴AM=AN,
∵∠FMA=∠BNA=90°,∴∠DGA=∠BGA,„„„„„„„„„10分
②3BEGE „„„„„„„„„„„„12分 理由：连接EF,由题意可知AE垂直平分BF,所以EF＝BE,又因∠EGF=120°,∠DGA=∠BGA（已证）
所以∠BGA＝60°,
第 9 页 共 9 页
由条件又可证EF⊥AC于点H,可得EH=FH,在Rt△GEH中
sinEHAGEGE
,即sin60EH
GE
 所以23EHGE,即3BEGE