构造三角形DEC,使其与三角形ABD全等,DE=BD=10, 取DE中点F,连BF,FC,BDF是直角三角形,计算得BF=5sqrt(5),FC=5,
三角形BFC中,BC<=BF+FC=5+5sqrt(5),等号在F在BC上时成立,这时,BC取得最大值5+5sqrt(5)

构造三角形DEC,使其与三角形ABD全等,DE=BD=10, 取DE中点F,连BF,FC,BDF是直角三角形,计算得BF=5sqrt(5),FC=5,
三角形BFC中,BC<=BF+FC=5+5sqrt(5),等号在F在BC上时成立,这时,BC取得最大值5+5sqrt(5)

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赞, 还是几何解法漂亮!
-kde235-
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12/07/2023 postreply
18:50:49
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