In quadrilateral ABCD,overrightarrow{AB}=vec{ | vedclass

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(B) In $	riangle ABC$,$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \vec{a} + \vec{b} \dots (i)$In $	riangle ADC$,$\overrightar

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are collinear

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(B) In $	riangle ABC$,$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \vec{a} + \vec{b} \dots (i)$In $	riangle ADC$,$\overrightarrow{AD} + \overrightarrow{DC} = \overrightarrow{AC}$. Given $\overrightarrow{DA} = \vec{a} - \vec{b}$,so $\overrightarrow{AD} = -(\vec{a} - \vec{b}) = \vec{b} - \vec{a}$.Thus,$\overrightarrow{DC} = \overrightarrow{AC} - \overrightarrow{AD} = (\vec{a} + \vec{b}) - (\vec{b} - \vec{a}) = 2\vec{a}$.$M$ is the midpoint of $BC$,so $\overrightarrow{BM} = \frac{1}{2}\overrightarrow{BC} = \frac{\vec{b}}{2}$.In $	riangle BDM$,$\overrightarrow{DM} = \overrightarrow{BM} - \overrightarrow{BD}$. Since $\overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD} = \vec{b} - 2\vec{a}$,we have $\overrightarrow{DM} = \frac{\vec{b}}{2} - (\vec{b} - 2\vec{a}) = 2\vec{a} - \frac{\vec{b}}{2} = \frac{4\vec{a} - \vec{b}}{2}$.Given $\overrightarrow{DX} = \frac{4}{5}\overrightarrow{DM} = \frac{4}{5} \left( \frac{4\vec{a} - \vec{b}}{2} ight) = \frac{8\vec{a} - 2\vec{b}}{5}$.Now,$\overrightarrow{AX} = \overrightarrow{AD} + \overrightarrow{DX} = (\vec{b} - \vec{a}) + \frac{8\vec{a} - 2\vec{b}}{5} = \frac{5\vec{b} - 5\vec{a} + 8\vec{a} - 2\vec{b}}{5} = \frac{3\vec{a} + 3\vec{b}}{5} = \frac{3}{5}(\vec{a} + \vec{b})$.Since $\overrightarrow{AX} = \frac{3}{5}\overrightarrow{AC}$,the vectors $\overrightarrow{AX}$ and $\overrightarrow{AC}$ are parallel and share a common point $A$.Therefore,the points $A, X$ and $C$ are collinear.

In quadrilateral $ABCD$,$\overrightarrow{AB}=\vec{a}$,$\overrightarrow{BC}=\vec{b}$,$\overrightarrow{DA}=\vec{a}-\vec{b}$. $M$ is the midpoint of $BC$ and $X$ is a point on $DM$ such that $\overrightarrow{DX}=\frac{4}{5} \overrightarrow{DM}$. Then the points $A, X$ and $C$:

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