Consider the parallelogram $ABCD$ and the points $M \in AB$ and $N \in AC$ such that:
$$\vec{AM} = \dfrac{1}{x} \vec{AB} \hspace{3cm} \vec{AN} = \dfrac{1}{y} \vec{AC}$$
with $x, y \in \mathbb{R}^{*}$. We have to find a relation between $x$ and $y$ such that the points $D, N, M$ are collinear.
I don't know how to approach this. The result given in my textbook is $x = y - 1$, but no explanation is given.
Let $\vec{AB}=\vec{a}$ and $\vec{AD}=\vec{b}.$
Thus, there is $k$ for which: $$\vec{MN}=k\vec{MB}$$ or $$-\frac{1}{x}\vec{a}+\frac{1}{y}(\vec{a}+\vec{b})=k\left(-\frac{1}{x}\vec{a}+\vec{b}\right),$$ which gives $$-\frac{1}{x}+\frac{1}{y}=-\frac{k}{x}$$ and $$\frac{1}{y}=k$$ and from here $$-\frac{1}{x}+\frac{1}{y}=-\frac{1}{xy}.$$ Can you end it now?