(2009•重庆)已知:如图,在直角梯形ABCD中,AD∥BC,∠ABC=90°,DE⊥AC于点F,交BC于点G,交AB的延长线于点E,且AE=AC.1.求证AE=AC2.若∠E=30,BC=3,求DC条件中不是AE=AC,是BG=FG,打错了= =
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![(2009•重庆)已知:如图,在直角梯形ABCD中,AD∥BC,∠ABC=90°,DE⊥AC于点F,交BC于点G,交AB的延长线于点E,且AE=AC.1.求证AE=AC2.若∠E=30,BC=3,求DC条件中不是AE=AC,是BG=FG,打错了= =](/uploads/image/z/8246641-49-1.jpg?t=%EF%BC%882009%26%238226%3B%E9%87%8D%E5%BA%86%EF%BC%89%E5%B7%B2%E7%9F%A5%EF%BC%9A%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E7%9B%B4%E8%A7%92%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAD%E2%88%A5BC%2C%E2%88%A0ABC%3D90%C2%B0%2CDE%E2%8A%A5AC%E4%BA%8E%E7%82%B9F%2C%E4%BA%A4BC%E4%BA%8E%E7%82%B9G%2C%E4%BA%A4AB%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%8E%E7%82%B9E%2C%E4%B8%94AE%3DAC%EF%BC%8E1.%E6%B1%82%E8%AF%81AE%3DAC2.%E8%8B%A5%E2%88%A0E%3D30%2CBC%3D3%2C%E6%B1%82DC%E6%9D%A1%E4%BB%B6%E4%B8%AD%E4%B8%8D%E6%98%AFAE%3DAC%2C%E6%98%AFBG%3DFG%EF%BC%8C%E6%89%93%E9%94%99%E4%BA%86%3D+%3D)
(2009•重庆)已知:如图,在直角梯形ABCD中,AD∥BC,∠ABC=90°,DE⊥AC于点F,交BC于点G,交AB的延长线于点E,且AE=AC.1.求证AE=AC2.若∠E=30,BC=3,求DC条件中不是AE=AC,是BG=FG,打错了= =
(2009•重庆)已知:如图,在直角梯形ABCD中,AD∥BC,∠ABC=90°,DE⊥AC于点F,交BC于点G,交AB的延长线于点E,且AE=AC.
1.求证AE=AC
2.若∠E=30,BC=3,求DC
条件中不是AE=AC,是BG=FG,打错了= =
(2009•重庆)已知:如图,在直角梯形ABCD中,AD∥BC,∠ABC=90°,DE⊥AC于点F,交BC于点G,交AB的延长线于点E,且AE=AC.1.求证AE=AC2.若∠E=30,BC=3,求DC条件中不是AE=AC,是BG=FG,打错了= =
你好,第一题是 求证:FC=BE 吧
证:∵∠DAC+∠CAE=90°,∠CAE+∠E=90°,
所以∠DAC=∠E
又AD∥BC,
所以∠DAC=∠ACB
所以∠E=∠ACB
又∠AFE=∠ABC=90°,AE=AC
所以△AEF全等于△ACB
所以AB=AF,
又AE=AC
故FC=BE
你好
第二题:
解得,AF=根号3,DF=1
CF=2根号3-根号3=根号3
所以,DC=2