The diagram shows the triangle \(OAC\). The point \(B\) lies on \(AC\) such that \(AB:BC=p:q\), where \(p\) and \(q\) are constants \((p\ne-q)\).
\(\overrightarrow{OA}=\mathbf a\), \(\overrightarrow{OB}=\mathbf b\) and \(\overrightarrow{OC}=\mathbf c\).
Show that \(\mathbf b=\dfrac{q\mathbf a+p\mathbf c}{q+p}\).[5]
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