In a parallelogram OABC with OA=vec a, OC=vec c point D divides OA in…

Let O be the initial point.

$\vec{OB}=\vec{OA}+\vec{AB}=\vec{OA}+\vec{OC}=\vec a+\vec c$

Also, $\vec{OD}=\frac{n}{n+1}\vec a$

Suppose OE : EB = λ : 1 and CE : ED = μ : 1.

Then $\vec{OE}=\frac{λ}{λ+1}.\vec{OB}=\frac{μ\vec{OD}+\vec{OC}}{μ+1}=\frac{λ}{λ+1}(\vec a+\vec c)=\frac{μ}{μ+1}.\frac{n}{n+1}.\vec a+\frac{1}{μ+1}.\vec c$

or $(\frac{λ}{λ+1}-\frac{μ}{μ+1}.\frac{n}{n+1})\vec a=(\frac{1}{μ+1}-\frac{λ}{λ+1})\vec c$

$⇒\frac{λ}{λ+1}-\frac{μ}{μ+1}.\frac{n}{n+1}=0$ and $\frac{1}{μ+1}-\frac{λ}{λ+1}=0$

(since $\vec a$ and $\vec c$ are non-collinear)

Hence solving for $λ, μ$, we find that $\frac{CE}{ED}=μ=\frac{n+1}{n}$

Hence (C) is the correct answer.