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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:BG=FG;
(2)若AD=DC=2,求AB的长.
人气:286 ℃ 时间:2020-10-01 12:28:08
解答
(1)证明:连接AG,∵∠ABC=90°,DE⊥AC于点F,∴∠ABC=∠AFE.在△ABC和△AFE中,∠ABC=∠AFE∠EAF=∠CABAC=AE∴△ABC≌△AFE(AAS),∴AB=AF.在Rt△ABG和Rt△AFG中,AG=AGAB=AF∴Rt△ABG≌Rt△AFG(HL)...
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