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