解道几何题在五边形ABCDE中,AB=BC=CD=DE,∠B=96°且∠C=∠D=108°.求∠E

Question

解道几何题
在五边形ABCDE中,AB = BC = CD = DE,∠B = 96° 且 ∠C = ∠D = 108°.求 ∠E

Answer 1

连接BD、CE交于点F
∵∠C为108°且BC=CD
∴∠CBD=∠CDB=36°
同理∠CED=∠ECD=36°
∵CD=DC ∠C=∠D BC=ED
∵△BCD≌△EDC
∴CE=BD
∵△BED≌△EBC
∴∠BED=∠EBC=72°
∴CD‖BE
∴∠EBD=∠CDB=36° ∠BEC=∠DCE=36°
∴BF=EF
∵∠BFE=108°
∴∠BFC=∠BCF=72°
∴BF=EF=BC=AB
∵∠ABF=∠ABC-∠FBC=96°-36°=60°
∴△ABC为等边三角形
∴∠AFB=60° 且AF=BF=EF
∴∠FAE=∠FEA,又知∠AFE=∠BFE-∠AFB=108°-60°=48°
∴∠AEF=(180°-48°)/2=66°
∴∠E=∠AEF+∠CED=102°