A vector along a trapezoid's diagonal 
In the trapezoid TXYZ, $\vec{TX} = 2 \vec{ZY},$ and the diagonals meet at $O.$ Find an expression for the vector $\vec{TO}$ in terms of the vectors $\vec{TX}$ and $\vec{TZ}.$

I do not know how to answer this question. It involves the properties of vectors.
What I have done so far:
Let $\vec{TX}$ = unit vector $\mathbf{a},$ then $\vec{ZY}=\frac12$ unit vector $\mathbf{a}.$
Could you please give me a hint?
 A: For a hint without reading the rest of the following answer, you could refer to this different-looking example, which utilises the same solution technique.
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We need to determine the position of point $O.$ Start by constructing a diagram, labelling it as helpfully as possible. As such, let
$$\vec{ZY}=\mathbf{a},\\
\vec{TZ}=\mathbf{b},\\
\vec{TO}=m\,\vec{TY},\\
\vec{ZO}=n\,\vec{ZX}.$$
Now we form an equation relating $m, n, \mathbf{a},$ and $\mathbf{b}:$

 $$\vec{TO}=\vec{TZ}+\vec{ZO}\\ m\,\vec{TY}=\mathbf{b}+n\,\vec{ZX}\\ m(\mathbf{a}+\mathbf{b})=\mathbf{b}+n(2\mathbf{a}- \mathbf{b})\\ (m-2n)\mathbf{a}+(m+n-1)\mathbf{b}=\mathbf{0}.$$

Since $\mathbf{a}$ and $\mathbf{b}$ are not collinear, the coefficients of both vectors are $0.$

 Therefore, $m=\frac23.$ So
 $$\vec{TO}=\frac23\vec{TY}\\ =\frac23(\mathbf{a}+\mathbf{b})\\ =\frac13(2\mathbf{a})+\frac23(\mathbf{b})\\ =\frac13\vec{TX}+\frac23\vec{TZ}.$$


Alternatively, to determine the position of point $O$ without relying just on vector properties, we can just use triangle similarity.

 Triangle $OTX$ is similar to triangle $OYZ.$ So
 $$\frac{OT}{OY}=\frac{TX}{YZ}\\=2.$$
 Thus $$TO=\frac23TY\\ \vec{TO}=\frac23\vec{TY}\\ =\frac23(\vec{TZ}+\vec{ZY})\\ =\frac23(\vec{TZ}+\frac12\vec{TX})\\ =\frac13\vec{TX}+\frac23\vec{TZ}.$$

