Point of intersection of two lines given colinearity

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I've a question regarding the following situation :

Given (for instance) a triangle $ABC$ and two points $M$ and $N$ on $(AB)$ and $(CD)$ respectively such that $(MN) \not \parallel (BC)$.

($k$ and $h$ are given such that $\vec{AM} = k \vec{AB}$ and $\vec{AN} = h\vec{AC}$)

If $J$ denotes the point of intersection of lines $(BC)$ and $(MN)$, it is asked to determine the real number $\alpha$ in terms of $k$ and $h$ such that

$$\vec{BJ} = \alpha \vec{BC}$$

My approach would be :

Determine the equations of the two lines $(MN)$ and $(BC)$ in the referential $(A, \vec{AB}, \vec{AC})$ and solve the system of equations to get the coordinates of $J$ first then do what is required to get the $\alpha$.

I wonder whether there is a better / shorter approach to tackle this problem.

Thanks.

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The answer depends on what is the information given, what lengths to write $\alpha$ from.

Menelaus' theorem allows you to express $\alpha$ as a function of the ratios that $M$ and $N$ divide the other sides of the triangle, and the side $BC$.