如图所示,已知AD与BC相交于点E,角1=角2=角3,BD=CD,角ADB=90度,CH垂直AB于点H,CH交AD于点F. (1)求证: CD平行AB. (2)求证:角BDE全等于角 ACE. (3)若O为AB的中点,求证:OF=二分之一BE.
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![如图所示,已知AD与BC相交于点E,角1=角2=角3,BD=CD,角ADB=90度,CH垂直AB于点H,CH交AD于点F. (1)求证: CD平行AB. (2)求证:角BDE全等于角 ACE. (3)若O为AB的中点,求证:OF=二分之一BE.](/uploads/image/z/14268333-21-3.jpg?t=%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%2C%E5%B7%B2%E7%9F%A5AD%E4%B8%8EBC%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9E%2C%E8%A7%921%3D%E8%A7%922%3D%E8%A7%923%2CBD%3DCD%2C%E8%A7%92ADB%3D90%E5%BA%A6%2CCH%E5%9E%82%E7%9B%B4AB%E4%BA%8E%E7%82%B9H%2CCH%E4%BA%A4AD%E4%BA%8E%E7%82%B9F.+++++++%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%3A+CD%E5%B9%B3%E8%A1%8CAB.+++%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%3A%E8%A7%92BDE%E5%85%A8%E7%AD%89%E4%BA%8E%E8%A7%92+ACE.++++%EF%BC%883%EF%BC%89%E8%8B%A5O%E4%B8%BAAB%E7%9A%84%E4%B8%AD%E7%82%B9%2C%E6%B1%82%E8%AF%81%3AOF%3D%E4%BA%8C%E5%88%86%E4%B9%8B%E4%B8%80BE.)
如图所示,已知AD与BC相交于点E,角1=角2=角3,BD=CD,角ADB=90度,CH垂直AB于点H,CH交AD于点F. (1)求证: CD平行AB. (2)求证:角BDE全等于角 ACE. (3)若O为AB的中点,求证:OF=二分之一BE.
如图所示,已知AD与BC相交于点E,角1=角2=角3,BD=CD,角ADB=90度,CH垂直AB于点H,CH交AD于点F. (1)求证: CD平行AB. (2)求证:角BDE全等于角 ACE. (3)若O为AB的中点,求证:OF=二分之一BE.
如图所示,已知AD与BC相交于点E,角1=角2=角3,BD=CD,角ADB=90度,CH垂直AB于点H,CH交AD于点F. (1)求证: CD平行AB. (2)求证:角BDE全等于角 ACE. (3)若O为AB的中点,求证:OF=二分之一BE.
证明:(1)∵BD=CD,
∴∠BCD=∠1;
∵∠1=∠2,
∴∠BCD=∠2;
∴CD∥AB.
(2)∵CD∥AB,∴∠CDA=∠3.
∵∠BCD=∠2=∠3,
∴BE=AE.
且∠CDA=∠BCD,
∴DE=CE.
在△BDE和△ACE中,
∵DE=CE,∠DEB=∠CEA,BE=AE.
∴△BDE≌△ACE;
(3)∵△BDE≌△ACE,
∴∠4=∠1,∠ACE=∠BDE=90°
∴∠ACH=90°-∠BCH;
又∵CH⊥AB,
∴∠2=90°-∠BCH;
∴∠ACH=∠2=∠1=∠4,
∴AF=CF;
∵∠AEC=90°-∠4,∠ECF=90°-∠ACH,
又∵∠ACH=∠4,
∴∠AEC=∠ECF;
∴CF=EF;
∴EF=AF;
∵O为AB中点,
∴OF为△ABE的中位线;
∴OF=1/2 BE.