如图,已知三角形ABC中角ACB等于90度,ACDE和BCFG都是正方形,过D,E,F,G,各点作直线AB的垂线,垂足分别为D1,E1,F1,G1,求证(1)AE1等于BG1;(2)EE1加GG1等于AB(3)DD1加FF1等于E1G1快
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![如图,已知三角形ABC中角ACB等于90度,ACDE和BCFG都是正方形,过D,E,F,G,各点作直线AB的垂线,垂足分别为D1,E1,F1,G1,求证(1)AE1等于BG1;(2)EE1加GG1等于AB(3)DD1加FF1等于E1G1快](/uploads/image/z/14294348-44-8.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%E8%A7%92ACB%E7%AD%89%E4%BA%8E90%E5%BA%A6%2CACDE%E5%92%8CBCFG%E9%83%BD%E6%98%AF%E6%AD%A3%E6%96%B9%E5%BD%A2%2C%E8%BF%87D%2CE%2CF%2CG%2C%E5%90%84%E7%82%B9%E4%BD%9C%E7%9B%B4%E7%BA%BFAB%E7%9A%84%E5%9E%82%E7%BA%BF%2C%E5%9E%82%E8%B6%B3%E5%88%86%E5%88%AB%E4%B8%BAD1%2CE1%2CF1%2CG1%2C%E6%B1%82%E8%AF%81%EF%BC%881%EF%BC%89AE1%E7%AD%89%E4%BA%8EBG1%EF%BC%9B%EF%BC%882%EF%BC%89EE1%E5%8A%A0GG1%E7%AD%89%E4%BA%8EAB%EF%BC%883%EF%BC%89DD1%E5%8A%A0FF1%E7%AD%89%E4%BA%8EE1G1%E5%BF%AB)
如图,已知三角形ABC中角ACB等于90度,ACDE和BCFG都是正方形,过D,E,F,G,各点作直线AB的垂线,垂足分别为D1,E1,F1,G1,求证(1)AE1等于BG1;(2)EE1加GG1等于AB(3)DD1加FF1等于E1G1快
如图,已知三角形ABC中角ACB等于90度,ACDE和BCFG都是正方形,过D,E,F,G,各点作直线AB的垂线,垂足分别为
D1,E1,F1,G1,求证(1)AE1等于BG1;(2)EE1加GG1等于AB(3)DD1加FF1等于E1G1
快
如图,已知三角形ABC中角ACB等于90度,ACDE和BCFG都是正方形,过D,E,F,G,各点作直线AB的垂线,垂足分别为D1,E1,F1,G1,求证(1)AE1等于BG1;(2)EE1加GG1等于AB(3)DD1加FF1等于E1G1快
证明:(1)过C点作CC1⊥AB,垂足为C1,则△AEE1≌△CAC1,△BGG1≌△CBC1,所以AE1=CC1=BG1;
(2)由(1)得EE1=AC1,GG1=BC1,所以EE1+GG1=AC1+BC1=AB;
(3)连DF,则△DCF≌△ACB,设DF交CC1的反向延长线于H,∠FDC=∠BAC=∠BCC1=∠HCD,所以DH=CH=HF,即H为AB的中点,则C1为D1F1的中点,所以DD1+FF1=2HC1=2(HC+CC1)=DF+2CC1=AB+AE1+BG1=E1G1.
求图 !!
求证:图呢???
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大哥图呢……
1)过C点作CC1⊥AB,垂足为C1,则△AEE1≌△CAC1,△BGG1≌△CBC1,所以AE1=CC1=BG1;
(2)由(1)得EE1=AC1,GG1=BC1,所以EE1+GG1=AC1+BC1=AB;
(3)连DF,则△DCF≌△ACB,设DF交CC1的反向延长线于H,∠FDC=∠BAC=∠BCC1=∠HCD,所以DH=CH=HF,即H为AB的中点,则C1为D1F1的中点,...
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1)过C点作CC1⊥AB,垂足为C1,则△AEE1≌△CAC1,△BGG1≌△CBC1,所以AE1=CC1=BG1;
(2)由(1)得EE1=AC1,GG1=BC1,所以EE1+GG1=AC1+BC1=AB;
(3)连DF,则△DCF≌△ACB,设DF交CC1的反向延长线于H,∠FDC=∠BAC=∠BCC1=∠HCD,所以DH=CH=HF,即H为AB的中点,则C1为D1F1的中点,所以DD1+FF1=2HC1=2(HC+CC1)=DF+2CC1=AB+AE1+BG1=E1G1.
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证明:(1)过C点作CC1⊥AB,垂足为C1,则△AEE1≌△CAC1,△BGG1≌△CBC1,所以AE1=CC1=BG1;
(2)由(1)得EE1=AC1,GG1=BC1,所以EE1+GG1=AC1+BC1=AB;
(3)连DF,则△DCF≌△ACB,设DF交CC1的反向延长线于H,∠FDC=∠BAC=∠BCC1=∠HCD,所以DH=CH=HF,即H为AB的中点,则C1为D1F1...
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证明:(1)过C点作CC1⊥AB,垂足为C1,则△AEE1≌△CAC1,△BGG1≌△CBC1,所以AE1=CC1=BG1;
(2)由(1)得EE1=AC1,GG1=BC1,所以EE1+GG1=AC1+BC1=AB;
(3)连DF,则△DCF≌△ACB,设DF交CC1的反向延长线于H,∠FDC=∠BAC=∠BCC1=∠HCD,所以DH=CH=HF,即H为AB的中点,则C1为D1F1的中点,所以DD1+FF1=2HC1=2(HC+CC1)=DF+2CC1=AB+AE1+BG1=E1G1.
(图请自己补充吧)
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