(1)①补全图形,如图1①.
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/80cb39dbb6fd526612d61e96a818972bd50736b6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
②连接MF,EF,如图1②.
∵AB=AC,∠BAC=60°,
∴△ABC是等边三角形,
∴CA=CB.
∵CE平分∠ACB,
∴CE⊥AB,即∠AEC=90°.
∵NF⊥CE,即∠FNC=90°,
∴∠AEC=∠FNC,
∴EH∥FN.
∵FH∥CE,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b90e7bec54e736d148b4ae8a98504fc2d4626988?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴四边形ENFH是平行四边形.
∵∠AEC=90°,
∴平行四边形ENFH是矩形.
∴EF=HN.
∵点M,N重合,
∴EF=HM.
故答案为:EF=HM.
(2)连接FM,如图2.
∵AD,CE分别平分∠BAC和∠ACB,且∠BAC=120°,
∴∠BAD=∠CAD=60°,∠ACE=∠BCE.
∵AB=AC,∴AD⊥BC.
∵NG⊥EC,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b8014a90f603738d45b72ea0b01bb051f919ecb7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴∠MDC=∠NGM=90°,
∴∠BCE+∠DMC=90°,∠MNG+∠DMC=90°.
∴∠BCE=∠MNG.
∴∠ACE=∠MNG.
∵NA=NC,∠NAC=60°,
∴△ANC是等边三角形,
∴AN=AC.
在△AFN和△AMC中,
,
∴△AFN≌△AMC(ASA),
∴AF=AM.
∴△AMF是等边三角形.
∴AF=FM,∠AMF=60°.
∴∠AMF=∠BAD.
∴FM∥AE.
∵FH∥CE,
∴四边形FHEM是平行四边形.
∴EH=FM.
∴AF=EH.
(3)连接BM,如图3.
∵AB=AC,∠BAC=36°,
∴∠ABC=∠ACB=72°.
∵CE平分∠ACB,
∴∠BCE=∠ACE=36°.
∵AB=AC,AD平分∠BAC,
∴AD垂直平分BC,∠BAD=18°,
∴MB=MC,NB=NC=AN,
∴∠MBC=∠MCD=36°,∠ABN=∠BAN=18°,
∴∠ABM=36°,∠BME=72°,∠NBC=72°-18°=54°,
∴∠BEM=72°=∠BME,∠NBC+∠ECD=54°+36°=90°,
∴BE=BM,BN⊥CE,
∴△BEM是黄金三角形.
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/e850352ac65c10381482ec40b1119313b17e89b7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴
=
.
∴EM=
BE.
∵NF⊥CE于点G,BN⊥CE,
∴B、G、N三点共线,
∴∠BGC=∠FGC=90°,即BG⊥EM.
∵BE=BM,BG⊥EM,
∴EG=MG=
EM=
BE.
在△BCG和△FCG中,
,
∴△BCG≌△FCG(ASA),
∴BG=FG.
∵EG∥FH,
∴
=
=1,
∴BE=EH=4,
∴MG=
BE=
-1.
∴MG的长为
-1.