在梯形ABCDABCDABCD中,AB//CD,AB=2CD,M,NAB//CD , AB=2CD , M , NAB//CD,AB=2CD,M,N分别是CD,BCCD , BCCD,BC的中点,若AB→=λAM→+μAN→\overrightarrow{AB}=\lambda\overrightarrow{AM}+\mu\overrightarrow{AN}AB=λAM+μAN,则 λ+μ=\lambda+\mu=λ+μ=___________。
【解析】
如图,延长 MNMNMN 交 ABABAB 的延长线与点 EEE,则有 △MNC≅△ENB\triangle MNC \cong \triangle ENB△MNC≅△ENB 则BEAE=14⇒AB→=45AE→\dfrac{BE}{AE}=\dfrac{1}{4}\Rightarrow \overrightarrow{AB}=\dfrac{4}{5}\overrightarrow{AE}AEBE=41⇒AB=54AE由 M,N,EM ,N ,EM,N,E 三点共线得:AN→=12AM→+12AE→\overrightarrow{AN}=\dfrac{1}{2}\overrightarrow{AM}+\dfrac{1}{2}\overrightarrow{AE}AN=21AM+21AE综合两式得:AB→=−45AM→+85AN→\overrightarrow{AB}=-\dfrac{4}{5}\overrightarrow{AM}+\dfrac{8}{5}\overrightarrow{AN}AB=−54AM+58AN则λ+μ=45.\lambda+\mu=\dfrac{4}{5}.λ+μ=54.