![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/7e3e6709c93d70cfe8ef76bcfbdcd100baa12b6e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:(1)①结论:AB∥DE,BC∥EF,CD∥AF.
证明:连接AD,如图1,
∵六边形ABCDEF是等角六边形,∴∠BAF=∠F=∠E=∠EDC=∠C=∠B=
=120°.
∵∠DAF+∠F+∠E+∠EDA=360°,∴∠DAF+∠EDA=360°-120°-120°=120°.
∵∠DAF+∠DAB=120°,∴∠DAB=∠EDA.∴AB∥DE.
同理BC∥EF,CD∥AF.
②结论:EF=BC,AF=DC.
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b7003af33a87e95047392fdd13385343fbf2b451?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
证明:连接AE、DB,如图2,
∵AB∥DE,AB=DE,∴四边形ABDE是平行四边形.
∴AE=DB,∠EAB=∠BDE.
∵∠BAF=∠EDC.∴∠FAE=∠CDB.
在△AFE和△DCB中,
.
∴△AFE≌△DCB.
∴EF=BC,AF=DC.
③结论:AB=DE,AF=DC,EF=BC.
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/c75c10385343fbf2c5027a92b37eca8065388f51?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
延长FE、CD相交于点P,延长EF、BA相交于点Q,延长DC、AB相交于点S,如图3.
∵六边形ABCDEF是等角六边形,∴∠BAF=∠AFE=120°.∴∠QAF=∠QFA=60°.
∴△QAF是等边三角形.∴∠Q=60°,QA=QF=AF.
同理:∠S=60°,SB=SC=BC;∠P=60°,PE=PD=ED.
∵∠S=∠P=60°,∴△PSQ是等边三角形.∴PQ=QS=SP.
∴QB=QS-BS=PS-CS=PC.∴AB+AF=AB+QA=QB=PC=PD+DC=ED+DC.
∵AB∥ED,∴△AOB~△DOE.∴
==.
同理:
=,
=.
∴
==.
∴
===
=1.
∴AB=ED,AF=DC,EF=BC.
(2)连接BF,如图4,
∵BC∥EF,∴∠CBF+∠EFB=180°.
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b64543a98226cffc0a036303ba014a90f603ea52?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∵∠A+∠ABF+∠AFB=180°,∴∠ABC+∠A+∠AFE=360°.
同理:∠A+∠ABC+∠C=360°.
∴∠AFE=∠C.
同理:∠A=∠D,∠ABC=∠E.
Ⅰ.若只有1个内角等于120°,不能保证该六边形一定是等角六边形.
反例:当∠A=120°,∠ABC=150°时,∠D=∠A∠=120°,∠E=∠ABC=150°.
∵六边形的内角和为720°,∴∠AFE=∠C=
(720°-120°-120°-150°-150°)=90°.
此时该六边形不是等角六边形.
Ⅱ.若有2个内角等于120°,也不能保证该六边形一定是等角六边形.
反例:当∠A=∠D=120°,∠ABC=150°时,∠E=∠ABC=150°.
∵六边形的内角和为720°,∴∠AFE=∠C=