In quadrilateral A B C D ,   A B → = a → ,   B C → = b → ,   D A…

In $∆ABC$,

$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$

$\Rightarrow \overrightarrow{AC}=\overrightarrow{a}+\overrightarrow{b} ...\left(i\right)$

In $∆ADC$,

$\overrightarrow{AD}+\overrightarrow{DC}=\overrightarrow{AC}$

$\Rightarrow \overrightarrow{DC}=\overrightarrow{AC}-\overrightarrow{AD}$

$\Rightarrow \overrightarrow{DC}=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{a}-\overrightarrow{b}=2\overrightarrow{a}$

$M$ is the midpoint of $BC$, so $\overrightarrow{MC}=\frac{\overrightarrow{b}}{2}$.

In $∆MDC$

$\overrightarrow{MD}+\overrightarrow{DC}=\overrightarrow{MC}$

$\Rightarrow \overrightarrow{MD}=\frac{\overrightarrow{b}}{2}-\overrightarrow{2a}$

$=\frac{\overrightarrow{b}-4\overrightarrow{a}}{2}$

$\Rightarrow \overrightarrow{DM}=-\overrightarrow{MD}=\frac{4\overrightarrow{a}-\overrightarrow{b}}{2}$

$X$ is a point on $DM$ such that $\overrightarrow{DX}=\frac{4}{5}\overrightarrow{DM}$.

In $∆ADX$

$\overrightarrow{AD}+\overrightarrow{DX}=\overrightarrow{AX}$

$\overrightarrow{AD}+\frac{4}{5}\overrightarrow{DM}=\overrightarrow{AX}$

$\overrightarrow{AX}=\overrightarrow{b}-\overrightarrow{a}+\frac{4}{5}\left(\frac{4\overrightarrow{a}-\overrightarrow{b}}{2}\right)=\left(\frac{16}{10}-1\right)\overrightarrow{a}+\left(1-\frac{4}{10}\right)\overrightarrow{b}$

$=\frac{3}{5}\left(\overrightarrow{a}+\overrightarrow{b}\right)$

Hence, $\overrightarrow{AX}=\frac{3}{5}\overrightarrow{AC}$ from $\left(i\right)$

i.e. the points $A, X$ and $C$ are collinear