一到希望杯试题 As shown in the figure 25,i
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一到希望杯试题
As shown in the figure 25,in△ABC we have AB=BC=CA,DF⊥AC,DE⊥AB.Please compare themagnitudes of the perimeter of △AEF and the perimeter of quadrilataral EBCF
中文大意: 在正△ABC中 AB=BC=CA,BC上有一点D,DE⊥AB,DF⊥AC.试比较三角形AEF和四边形EBCF的周长
![](http://img.wesiedu.com/upload/4/3d/43d15442ea65a6affd68e913c52cf9d9.jpg)
As shown in the figure 25,in△ABC we have AB=BC=CA,DF⊥AC,DE⊥AB.Please compare themagnitudes of the perimeter of △AEF and the perimeter of quadrilataral EBCF
中文大意: 在正△ABC中 AB=BC=CA,BC上有一点D,DE⊥AB,DF⊥AC.试比较三角形AEF和四边形EBCF的周长
![](http://img.wesiedu.com/upload/4/3d/43d15442ea65a6affd68e913c52cf9d9.jpg)
![一到希望杯试题 As shown in the figure 25,i](/uploads/image/z/17094795-51-5.jpg?t=%E4%B8%80%E5%88%B0%E5%B8%8C%E6%9C%9B%E6%9D%AF%E8%AF%95%E9%A2%98+As%26nbsp%3Bshown%26nbsp%3Bin%26nbsp%3Bthe%26nbsp%3Bfigure%26nbsp%3B25%2Ci)
因为AB=BC=CA,且ED⊥AB,DF⊥AC,则∠EDB=∠FDC=30°,所以BD=2BE,DC=2FC,则四边形BCFE=EF+BE+FC+BD+DC=3BE+3FC+EF,
而C△AEF=AE+EF+AF=AB-BE+AC-FC+EF=2BC-BE-FC+EF=2(BD+DC)-BE-FC+EF=3BE+3FC+EF.
所以三角形AEF和四边形EBCF的周长相等
而C△AEF=AE+EF+AF=AB-BE+AC-FC+EF=2BC-BE-FC+EF=2(BD+DC)-BE-FC+EF=3BE+3FC+EF.
所以三角形AEF和四边形EBCF的周长相等