在平行四边形ABCD中,E为BC边上一点,联结ED,且∠AED=∠B,在DE上取一点F,使AF=AE(1)请写出图中所有相似的三角形,并证明你结论的正确性.(2)若AE=2√3,BC=3BE,求DE*DF的值.3Q了!如图
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![在平行四边形ABCD中,E为BC边上一点,联结ED,且∠AED=∠B,在DE上取一点F,使AF=AE(1)请写出图中所有相似的三角形,并证明你结论的正确性.(2)若AE=2√3,BC=3BE,求DE*DF的值.3Q了!如图](/uploads/image/z/8639884-28-4.jpg?t=%E5%9C%A8%E5%B9%B3%E8%A1%8C%E5%9B%9B%E8%BE%B9%E5%BD%A2ABCD%E4%B8%AD%2CE%E4%B8%BABC%E8%BE%B9%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E8%81%94%E7%BB%93ED%2C%E4%B8%94%E2%88%A0AED%3D%E2%88%A0B%2C%E5%9C%A8DE%E4%B8%8A%E5%8F%96%E4%B8%80%E7%82%B9F%2C%E4%BD%BFAF%3DAE%EF%BC%881%EF%BC%89%E8%AF%B7%E5%86%99%E5%87%BA%E5%9B%BE%E4%B8%AD%E6%89%80%E6%9C%89%E7%9B%B8%E4%BC%BC%E7%9A%84%E4%B8%89%E8%A7%92%E5%BD%A2%2C%E5%B9%B6%E8%AF%81%E6%98%8E%E4%BD%A0%E7%BB%93%E8%AE%BA%E7%9A%84%E6%AD%A3%E7%A1%AE%E6%80%A7.%EF%BC%882%EF%BC%89%E8%8B%A5AE%3D2%E2%88%9A3%2CBC%3D3BE%2C%E6%B1%82DE%2ADF%E7%9A%84%E5%80%BC.3Q%E4%BA%86%21%E5%A6%82%E5%9B%BE)
在平行四边形ABCD中,E为BC边上一点,联结ED,且∠AED=∠B,在DE上取一点F,使AF=AE(1)请写出图中所有相似的三角形,并证明你结论的正确性.(2)若AE=2√3,BC=3BE,求DE*DF的值.3Q了!如图
在平行四边形ABCD中,E为BC边上一点,联结ED,且∠AED=∠B,在DE上取一点F,使AF=AE
(1)请写出图中所有相似的三角形,并证明你结论的正确性.
(2)若AE=2√3,BC=3BE,求DE*DF的值.
3Q了!
如图
在平行四边形ABCD中,E为BC边上一点,联结ED,且∠AED=∠B,在DE上取一点F,使AF=AE(1)请写出图中所有相似的三角形,并证明你结论的正确性.(2)若AE=2√3,BC=3BE,求DE*DF的值.3Q了!如图
(1)△ABE∽△DEA,△AFD∽△DCE.
(2)∵BC=3BE,
∴设BE=x,则BC=3x,
∴AD=3x,EC=2x,
由△ABE∽△DEA,得: AEAD=BEAE,
∵ AE=23,
∴ 233x=x23,
∴x=2,
又由△AFD∽△DCE,
得DE•DF=AD•EC=3x×2x=6x2,
∴DE•DF=24.
故答案为:24.
(1)△ABE∽△DEA,△ADF∽△DEC,
证明:∵∠B=∠AED,∠AEB=∠DAE=90°,∴△ABE∽△DEA;
∵在平行四边形中,∴∠ADE=∠DEC,∠B+∠DCE=180°=∠AEF+∠DCE=180°
又 AE=AF,∴∠AEF=∠AFE,
∵∠AFE+∠AFD=180°,∴∠AFD=∠DCE,∴△ADF∽△DEC
(2)BC=3BE,设B...
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(1)△ABE∽△DEA,△ADF∽△DEC,
证明:∵∠B=∠AED,∠AEB=∠DAE=90°,∴△ABE∽△DEA;
∵在平行四边形中,∴∠ADE=∠DEC,∠B+∠DCE=180°=∠AEF+∠DCE=180°
又 AE=AF,∴∠AEF=∠AFE,
∵∠AFE+∠AFD=180°,∴∠AFD=∠DCE,∴△ADF∽△DEC
(2)BC=3BE,设BE=x,则BC=AD=3x,EC=2x,
由△ABE∽△DEA,得:AE/AD=BE/AE,∵AE=2√3,∴2√3/3x=x/2√3,∴x=2
又由△ADF∽△DEC,得:DF/EC=AD/DE,即DF×DE=AD×EC=3x·2x=3×2×2×2=24
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(1)△ABE∽△DEA,△ADF∽△DEC,
证明:∵∠B=∠AED,∠AEB=∠DAE=90°,∴△ABE∽△DEA;
∵在平行四边形中,∴∠ADE=∠DEC,∠B+∠DCE=180°=∠AEF+∠DCE=180°
又 AE=AF,∴∠AEF=∠AFE,
∵∠AFE+∠AFD=180°,∴∠AFD=∠DCE,∴△ADF∽△DEC
(2)BC=3BE,设B...
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(1)△ABE∽△DEA,△ADF∽△DEC,
证明:∵∠B=∠AED,∠AEB=∠DAE=90°,∴△ABE∽△DEA;
∵在平行四边形中,∴∠ADE=∠DEC,∠B+∠DCE=180°=∠AEF+∠DCE=180°
又 AE=AF,∴∠AEF=∠AFE,
∵∠AFE+∠AFD=180°,∴∠AFD=∠DCE,∴△ADF∽△DEC
(2)BC=3BE,设BE=x,则BC=AD=3x,EC=2x,
由△ABE∽△DEA,得:AE/AD=BE/AE,∵AE=2√3,∴2√3/3x=x/2√3,∴x=2
又由△ADF∽△DEC,得:DF/EC=AD/DE,即DF×DE=AD×EC=3x·2x=3×2×2×2=24
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