题目内容
把两个全等的直角三角板ABC和EFG叠放在一起,且使三角板EFG的直角顶点G与三角板ABC的斜边中点O重合,其中∠B=∠F=30°,斜边AB和EF的长均为4。
(1)当EG⊥AC于点K,GF⊥BC于点H时,如图23-1,求GH:GK的值.
(2)现将三角板EFG由图23-1所示的位置绕O点沿逆时针方向旋转,旋转
角满足条件:
0°<
<30°,如图23-2,EG交AC于点K,GF交BC于点H,GH:GK的值是否改变?证明你的结论.
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(1)当EG⊥AC于点K,GF⊥BC于点H时,如图23-1,求GH:GK的值.
(2)现将三角板EFG由图23-1所示的位置绕O点沿逆时针方向旋转,旋转
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0°<
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(1)
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(2)证明略
(1)解:∵GE⊥AC于 K,GF⊥BC于H,
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∴∠AKG =∠GHB =90°
∵∠ACB =90°
∴GK∥BC……………………………(1分)
∴∠AGK =∠B =30°………………(2分)
∵G与AB的中点O重合
∴AG = GB
∴△AKG≌△GHB……………………(3分)
∴KG = HB……………………………(4分)
在Rt△GHB中,tan∠B =
…(5分)
∴
……………………………(6分)
(2)GH:GK的值不改变。………………………(7分)
证明:过点G作GP⊥AC于点P,GQ⊥BC于点Q,
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∵∠C = 90°
∴四边形PCQG是矩形……………………(8分)
∴∠PGK+∠KGO = 90°
∵∠EGF = 90°
∴∠HGQ+∠KGQ = 90°
∴∠PGK = ∠HGQ ………………………(9分)
∵∠GPK =∠GQH = 90°
∴△PGK∽△QGH…………………………(10分)
∴
由(1)可得:
…… (11分)
∴
………………………………(12分)
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∴∠AKG =∠GHB =90°
∵∠ACB =90°
∴GK∥BC……………………………(1分)
∴∠AGK =∠B =30°………………(2分)
∵G与AB的中点O重合
∴AG = GB
∴△AKG≌△GHB……………………(3分)
∴KG = HB……………………………(4分)
在Rt△GHB中,tan∠B =

∴
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(2)GH:GK的值不改变。………………………(7分)
证明:过点G作GP⊥AC于点P,GQ⊥BC于点Q,

∵∠C = 90°
∴四边形PCQG是矩形……………………(8分)
∴∠PGK+∠KGO = 90°
∵∠EGF = 90°
∴∠HGQ+∠KGQ = 90°
∴∠PGK = ∠HGQ ………………………(9分)
∵∠GPK =∠GQH = 90°
∴△PGK∽△QGH…………………………(10分)
∴
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∴
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