在正方形ABCD中,AB=4,E为CD上一动点,AE交BD于F,过F作FH⊥AE于H,过H作GH⊥BD于G在正方形ABCD中,AB=4,E为CD上一动点,AE交BD于F,过F作FH⊥AE于H,过H作GH⊥BD于G,下列有四个结论:①AF=FH,②∠HAE=45°,③BD=2F
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![在正方形ABCD中,AB=4,E为CD上一动点,AE交BD于F,过F作FH⊥AE于H,过H作GH⊥BD于G在正方形ABCD中,AB=4,E为CD上一动点,AE交BD于F,过F作FH⊥AE于H,过H作GH⊥BD于G,下列有四个结论:①AF=FH,②∠HAE=45°,③BD=2F](/uploads/image/z/5553161-17-1.jpg?t=%26%238203%3B%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%2CAB%3D4%2CE%E4%B8%BACD%E4%B8%8A%E4%B8%80%E5%8A%A8%E7%82%B9%2CAE%E4%BA%A4BD%E4%BA%8EF%2C%E8%BF%87F%E4%BD%9CFH%E2%8A%A5AE%E4%BA%8EH%2C%E8%BF%87H%E4%BD%9CGH%E2%8A%A5BD%E4%BA%8EG%E5%9C%A8%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%E4%B8%AD%2CAB%3D4%2CE%E4%B8%BACD%E4%B8%8A%E4%B8%80%E5%8A%A8%E7%82%B9%2CAE%E4%BA%A4BD%E4%BA%8EF%2C%E8%BF%87F%E4%BD%9CFH%E2%8A%A5AE%E4%BA%8EH%2C%E8%BF%87H%E4%BD%9CGH%E2%8A%A5BD%E4%BA%8EG%2C%E4%B8%8B%E5%88%97%E6%9C%89%E5%9B%9B%E4%B8%AA%E7%BB%93%E8%AE%BA%EF%BC%9A%E2%91%A0AF%3DFH%2C%E2%91%A1%E2%88%A0HAE%3D45%C2%B0%2C%E2%91%A2BD%3D2F)
在正方形ABCD中,AB=4,E为CD上一动点,AE交BD于F,过F作FH⊥AE于H,过H作GH⊥BD于G在正方形ABCD中,AB=4,E为CD上一动点,AE交BD于F,过F作FH⊥AE于H,过H作GH⊥BD于G,下列有四个结论:①AF=FH,②∠HAE=45°,③BD=2F
在正方形ABCD中,AB=4,E为CD上一动点,AE交BD于F,过F作FH⊥AE于H,过H作GH⊥BD于G
在正方形ABCD中,AB=4,E为CD上一动点,AE交BD于F,过F作FH⊥AE于H,过H作GH⊥BD于G,下列有四个结论:①AF=FH,②∠HAE=45°,③BD=2FG,④△CEH的周长为定值,其中正确的结论有( )第四个中△MEC≌△MIC为什么?
延长AD至点M,使AD=DM,过点C作CI∥HL,则:LI=HC,
根据△MEC≌△CIM,可得:CE=IM,
同理,可得:AL=HE,
∴HE+HC+EC=AL+LI+IM=AM=8.
∴△CEH的周长为8,为定值.
在正方形ABCD中,AB=4,E为CD上一动点,AE交BD于F,过F作FH⊥AE于H,过H作GH⊥BD于G在正方形ABCD中,AB=4,E为CD上一动点,AE交BD于F,过F作FH⊥AE于H,过H作GH⊥BD于G,下列有四个结论:①AF=FH,②∠HAE=45°,③BD=2F
满意回答:
(1)连接FC,延长HF交AD于点L,
∵BD为正方形ABCD的对角线,
∴∠ADB=∠CDF=45°.
∵AD=CD,DF=DF,
∴△ADF≌△CDF.
∴FC=AF,∠ECF=∠DAF.
∵∠ALF+∠LAF=90°,
∴∠LHC+∠DAF=90°.
∵∠ECF=∠DAF,
∴∠FHC=∠FCH,
∴FH=FC.
∴FH=AF.
∵FH⊥AE,FH=AF,
∴∠HAE=45°.
(2)连接AC交BD于点O,可知:BD=2OA,
∵∠AFO+∠GFH=∠GHF+∠GFH,
∴∠AFO=∠GHF.
∵AF=HF,∠AOF=∠FGH=90°,
∴△AOF≌△FGH.
∴OA=GF.
∵BD=2OA,
∴BD=2FG.
(3)延长AD至点M,使AD=DM,过点C作CI∥HL,则:LI=HC,
根据△MEC≌△MIC,可得:CE=IM,
同理,可得:AL=HE,
∴HE+HC+EC=AL+LI+IM=AM=8.
∴△CEH的周长为8,为定值.