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九年级数学解答题一般
题目
如图,在梯形ABCDABCD,AB,ABCDCD,EECDCD的中点,且EC=13AB,ACEC=\frac{1}{3}AB,ACBEBE交于点FF.
(1)(1)AB=m,AD=n\overrightarrow{AB}=\overrightarrow{m},\overrightarrow{AD}=\overrightarrow{n},请用m,n\overrightarrow{m},\overrightarrow{n}来表示DCAF\overrightarrow{DC}、\overrightarrow{AF}
(2)(2)在原图中直接在图中作出AC\overrightarrow{AC}m,n\overrightarrow{m},\overrightarrow{n}方向上的分向量(不要求写作法,但要写出所作图中表示结论的向量).
知识点:梯形的定义、平面向量I章节:第22章 四边形 / 第4节 平面向量及其加减运算 / 22.7 平面向量

答案与解析

答案

(1)AB\left(1\right)\because ABCDCDEC=13ABEC=\frac{1}{3}AB
EC=13m\therefore \overrightarrow{EC}=\frac{1}{3}\overrightarrow{m}.
E\because ECDCD的中点,
DC=2EC=23m\therefore \overrightarrow{DC}=2\overrightarrow{EC}=\frac{2}{3}\overrightarrow{m}.
EC\because ECABAB
CFAF=ECAB=13\therefore \frac{CF}{AF}=\frac{EC}{AB}=\frac{1}{3}
AF=34AC\therefore AF=\frac{3}{4}AC
AC=AD+DC=n+23m\because \overrightarrow{AC}=\overrightarrow{AD}+\overrightarrow{DC}=\overrightarrow{n}+\frac{2}{3}\overrightarrow{m}
AF=34n+12m\therefore \overrightarrow{AF}=\frac{3}{4}\overrightarrow{n}+\frac{1}{2}\overrightarrow{m}.
(2)(2)如图,过点CCCTCTADADABAB于点TT
AD\overrightarrow{AD}AT\overrightarrow{AT}即为所求.

解析

(1)AB\left(1\right)\because ABCDCDEC=13ABEC=\frac{1}{3}AB
EC=13m\therefore \overrightarrow{EC}=\frac{1}{3}\overrightarrow{m}.
E\because ECDCD的中点,
DC=2EC=23m\therefore \overrightarrow{DC}=2\overrightarrow{EC}=\frac{2}{3}\overrightarrow{m}.
EC\because ECABAB
CFAF=ECAB=13\therefore \frac{CF}{AF}=\frac{EC}{AB}=\frac{1}{3}
AF=34AC\therefore AF=\frac{3}{4}AC
AC=AD+DC=n+23m\because \overrightarrow{AC}=\overrightarrow{AD}+\overrightarrow{DC}=\overrightarrow{n}+\frac{2}{3}\overrightarrow{m}
AF=34n+12m\therefore \overrightarrow{AF}=\frac{3}{4}\overrightarrow{n}+\frac{1}{2}\overrightarrow{m}.
(2)(2)如图,过点CCCTCTADADABAB于点TT
AD\overrightarrow{AD}AT\overrightarrow{AT}即为所求.

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