$解:如圖,E為AC中點(diǎn),F為BC中點(diǎn)$ $①:當(dāng)C在AB之間時(shí),EF=EC+CF=\frac {1}{2}AC+\frac {1}{2}CB=4$ $②:當(dāng)B在AC之間時(shí),EF=EC-CF=\frac {1}{2}AC-\frac {1}{2}CB=1$ 綜上,它們距離為1或4 解:∵D是BC中點(diǎn),∴DC=DB ∴AB=AC+BC=AC+2BD=8 解:是,理由: ∵AD=BC,即AC+CD=BD+CD ∴AC=BD ∵E是AB中點(diǎn),∴AE=BE ∴AC+CE=BD+DE ∴CE=AE-AC,DE=BE-BD 又∵AE=BE,AC=BD ∴CE=DE,即E為CD中點(diǎn) $(2)解:∵D,E分別是線段AC,BC中點(diǎn)$ $∴DC=\frac {1}{2}AC,CE=\frac {1}{2}CB$ $∴DE=DC+CE=\frac {1}{2}(AC+CB)=\frac {1}{2}AB=6$
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