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2(2)は△DFEと△BCFが相似の関係なので
DE:BF＝EF:FCで求められます
3はAD:a AE＝1:2よりEC＝2a
同様にBE:BD＝1:2よりDG＝4a
FG＝DG−DFなので4a−a＝3a
です！
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