P is any point on the segment $\stackrel{\rightharpoonup }{BC}$ of $∆$ABC. If Q is a point such that $\stackrel{\rightharpoonup }{PQ}$=$\stackrel{\rightharpoonup }{AP}+\stackrel{\rightharpoonup }{PB}+\stackrel{\rightharpoonup }{PC}$, then $\stackrel{\rightharpoonup }{AQ}$ is equal to

$\stackrel{\rightharpoonup }{AP}+\stackrel{\rightharpoonup }{PB}+\stackrel{\rightharpoonup }{PC}=\stackrel{\rightharpoonup }{PQ}$

$\Rightarrow$$\stackrel{\rightharpoonup }{AB}+\stackrel{\rightharpoonup }{PC}=\stackrel{\rightharpoonup }{PQ}\Rightarrow \stackrel{\rightharpoonup }{AB}=\stackrel{\rightharpoonup }{PQ}-\stackrel{\rightharpoonup }{PC}$

Also $\stackrel{\rightharpoonup }{CQ}=\stackrel{\rightharpoonup }{PQ}-\stackrel{\rightharpoonup }{PC}$

$\therefore$$\stackrel{\rightharpoonup }{AB}=\stackrel{\rightharpoonup }{CQ}$$\stackrel{ }{\to }$(1)

Similarly $\stackrel{\rightharpoonup }{AP}+\stackrel{\rightharpoonup }{PB}+\stackrel{\rightharpoonup }{PC}=\stackrel{\rightharpoonup }{AC}+\stackrel{\rightharpoonup }{PD}=\stackrel{\rightharpoonup }{PQ}$

$\Rightarrow$$\stackrel{\rightharpoonup }{AC}=\stackrel{\rightharpoonup }{PQ}-\stackrel{\rightharpoonup }{PB}=\stackrel{\rightharpoonup }{BQ}$$\underset{ }{\to }$(2)

(1) and (2)$\Rightarrow$$\stackrel{\rightharpoonup }{CQ}=\stackrel{\rightharpoonup }{BQ}$

Now $\stackrel{\rightharpoonup }{AQ}=\stackrel{\rightharpoonup }{AB}+\stackrel{\rightharpoonup }{BQ}=\stackrel{\rightharpoonup }{AB}+\stackrel{\rightharpoonup }{AC}$