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電子課本網(wǎng) 第63頁

第63頁

信息發(fā)布者:
$27^{\circ}$
$4\sqrt{3}$
$12$
解:?$(1)$?∵五邊形?$ABCDE$?是正五邊形,
∴?$∠ABC=\frac {(5 - 2)×180°}{5}=108°。$?
?$ (2) \triangle AMN$?是正三角形。理由如下:
?$ $?連接?$ON、$??$NF。$?由作圖,得?$FN = OF。$?
∵?$OF = ON,$
?∴?$FN = OF = ON,$?
∴?$\triangle FON$?是等邊三角形,
∴?$∠NFA = 60°。$?
∵?$\overset {\frown }{AN}=\overset {\frown }{AN},$?
∴?$∠NMA=∠NFA = 60°。$?
同理,可得?$∠ANM = 60°。$?
?$ $?在?$\triangle AMN$?中,?$∠MAN = 60°,$?
∴?$∠NMA=∠ANM=∠MAN,$?
∴?$\triangle AMN$?是正三角形。
?$ (3)$?由?$(2),$?得?$\triangle AMN$?是正三角形,
∴?$∠AON = 2∠AMN = 120°,$?
∴?$\overset {\frown }{AN}=120°。$?
∵?$\overset {\frown }{AD}=2\overset {\frown }{AE}=2×\frac {360°}{5}=144°,$?
∴?$\overset {\frown }{DN}=\overset {\frown }{AD}-\overset {\frown }{AN}=144°-120°=24°,$?
∴?$n=\frac {360°}{24°} = 15。$?
證明:$∵\(yùn)triangle ABC$是等邊三角形,
$∴∠A=∠B=∠C = 60°,$$AB = BC = AC。$
$∵E$、$F$、$G$、$H$、$L$、$K$分別是各邊的三等分點(diǎn),
$∴AE = EF = FB=\frac{1}{3}AB,$$BG = GH = HC=\frac{1}{3}BC,$$LC = KL = AK=\frac{1}{3}AC,$
$∴AE = AK,$$BF = BG,$$CH = CL。$
$∵∠A=∠B=∠C = 60°,$
$∴\triangle AEK$、$\triangle BGF$、$\triangle CHL$是等邊三角形,
$∴EK = EF = FG = GH = HL = LK,$
$∠KEF=∠EFG=∠FGH=∠GHL=∠HLK=∠LKE = 120°,$
$∴$六邊形$EFGHLK$是正六邊形。