已知边长为3的正方形ABCD,点E在射线BC上且BE=2CE,连接AE交射线DC于点F,设H在射线CD上使角EAH=∠BAE1,如图1当点E在线段BC上,求CF的长2,求sin∠DAH
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![已知边长为3的正方形ABCD,点E在射线BC上且BE=2CE,连接AE交射线DC于点F,设H在射线CD上使角EAH=∠BAE1,如图1当点E在线段BC上,求CF的长2,求sin∠DAH](/uploads/image/z/6138758-38-8.jpg?t=%E5%B7%B2%E7%9F%A5%E8%BE%B9%E9%95%BF%E4%B8%BA3%E7%9A%84%E6%AD%A3%E6%96%B9%E5%BD%A2ABCD%2C%E7%82%B9E%E5%9C%A8%E5%B0%84%E7%BA%BFBC%E4%B8%8A%E4%B8%94BE%3D2CE%2C%E8%BF%9E%E6%8E%A5AE%E4%BA%A4%E5%B0%84%E7%BA%BFDC%E4%BA%8E%E7%82%B9F%2C%E8%AE%BEH%E5%9C%A8%E5%B0%84%E7%BA%BFCD%E4%B8%8A%E4%BD%BF%E8%A7%92EAH%3D%E2%88%A0BAE1%2C%E5%A6%82%E5%9B%BE1%E5%BD%93%E7%82%B9E%E5%9C%A8%E7%BA%BF%E6%AE%B5BC%E4%B8%8A%2C%E6%B1%82CF%E7%9A%84%E9%95%BF2%2C%E6%B1%82sin%E2%88%A0DAH)
已知边长为3的正方形ABCD,点E在射线BC上且BE=2CE,连接AE交射线DC于点F,设H在射线CD上使角EAH=∠BAE1,如图1当点E在线段BC上,求CF的长2,求sin∠DAH
已知边长为3的正方形ABCD,点E在射线BC上且BE=2CE,连接AE交射线DC于点F,设H在射线CD上使角EAH=∠BAE
1,如图1当点E在线段BC上,求CF的长
2,求sin∠DAH
已知边长为3的正方形ABCD,点E在射线BC上且BE=2CE,连接AE交射线DC于点F,设H在射线CD上使角EAH=∠BAE1,如图1当点E在线段BC上,求CF的长2,求sin∠DAH
1.因为AD∥BC
∴△EAB∼△EFC
∴CF/AB=CE/BE=1/2
∴CF=AB/2=3/2
2.延长AB1交DC于H,因为∠BAE=∠B1AE=∠DFE
∴AH=FH,AE=√((3^2)+(2^2))=√(13)
FE/AE=CE/BE=1/2
∴EF=√(13)/2
∴AF=3√(13)/2
∴(BA^2)=AG•EA⇒(3^2)=AG•√(13)⇒AG=9/√(13)
易知△FKH∼△AGB
∴FH/3=(3√(13)/4)/(9/√(13))⇒FH=13/4
∴DH=3+1.5-3.25=5/4
∴AH=√((3^2)+((5/4)^2))=13/4
∴sin∠DAB1=DH/AH=(5/4)/(13/4)=5/13
当点E在BC延长线上时:过B1作B1H⊥BA延长线于H,
AE=√((3^2)+(6^2))=3√(5)
(AB^2)=AG•AE⇒(3^2)=AG•3√(5)⇒AG=3/√(5)
BG=(3•6)/(3√(5))=6/√(5)
∴BB1=12/√(5)
△B1HB∼△ABG
∴BH/BG=BB1/AB⇒BH/(6/√(5))=(12/√(5))/3⇒BH=24/5
∴AH=24/5-3=9/5
sin∠DAB1=sin∠HB1A=AH/AB1=(9/5)/3=3/5
3.当E点在线段BC上时
因为BE=X•CE CE=3-BE
∴BE=X•(3-BE)⇒BE=3X/(X+1)
∴S=y=AB•BE=3•3X/(X+1)
y=9x/(x+1)(x>0)
当E点在线段BC上时y=9x/2(x+1)(x>0)
当E点在线段BC延长线上时:
当E点在BC延长线上时:
BE=X•CE⇒BC+CE=X•CE⇒BC=(X-1)CE⇒3=(X-1)CE
∴CE=3/(X-1)
∴BE=BC+CE=3+[3/(X-1)]=3X/(X-1)
∴S△ABE=9X/2(X-1)
S△CEF/S△ABE=((CE/BE)^2)=1/(X^2)
∴S△CEF=[9X/2(X-1)](1/(X^2))=9/2X(X-1)
∴S△CEF/S△ADF=((CE/AD)^2)=({[3/(X-1)]/3}^2)
S△ADF=[9/2X(X-1)]/([1/(X-1)]^2)
∴y=[9(x-1)]/2x(x>1) 答案是我在另一个回答那里复制过来的