Euclidean Geometry Quadrilateral Problem 
When the quadrilateral is a square, rectangle, or parallelogram, the problem is very simple since X1Y1=X2Y2=X3Y3=a=b, but this falls apart when the quadrilateral is something like a trapezoid. How can I generalize this to any quadrilateral? This problem has me stumped. 
 A: ETA: The below assumes that it's $\mathbf{a}$ and $\mathbf{b}$ that are fixed, not $A, B,$ and $C$.  I may have tried to get too cute with this; perhaps the least confusing way to do this is to represent each of the points on the quadrilateral with $x$-$y$ coordinates ($A$ being at the origin, without loss of generality), and then express each of the desired vectors as a mix $\alpha\mathbf{a}+(1-\alpha)\mathbf{b}$.

Any vector doesn't really care where it starts, only what direction it's in.  Therefore, without loss of generality, identify $D$ with $A$; that is, in the triangle $ABC$, let $\vec{AB} = \mathbf{a}, \vec{AC} = \mathbf{b}$.  Divide $\overline{BC}$ into four equal parts with $Y_1, Y_2, Y_3$.  Then the three desired vectors are $\vec{AY_1}, \vec{AY_2}, \vec{AY_3}$.
To make this rigorous, you should prove that the above works.  There are a number of different ways to demonstrate it, but analytically on the Cartesian plane may be the most straightforward.
A: Arithmetic progression of vectors in the plane with a common vector difference $ \dfrac{b-a}{4}  $:
$$ 4 a /4 , (3 a +b)/4, (2 a + 2 b)/4 , ( a +3 b)/4 , 4 b /4. $$
If instead you had asked for $ \vec{AD}, \vec{BC}, $ then you get same quarter division intermediate grid points writing $ a \rightarrow c ,b \rightarrow d. $ 
A: $\vec{AB}+\vec{BC}+\vec{CD}+\vec{DA}=0$
$\vec{a}+\vec{BC}-\vec{b}+\vec{DA}=0$
$\vec{BC}+\vec{DA}=\vec{b}-\vec{a}$
$X_{1}Y_{1}=\frac{DA}{4}+a+\frac{BC}{4}=a+\frac{b-a}{4}=\frac{3a}{4}+\frac{b}{4}$
$X_{2}Y_{2}=\frac{DA}{2}+a+\frac{BC}{4}=a+\frac{(b-a)}{2}=\frac{a}{2}+\frac{b}{2}$
$X_{3}Y_{3}=\frac{3DA}{4}+a+\frac{BC}{4}=a+\frac{3(b-a)}{4}=\frac{a}{4}+\frac{3b}{4}$
A: Since
\begin{align}
X_i&=A\cdot(1-t_i)+D\cdot t_i,
\\
Y_i&=B\cdot(1-t_i)+C\cdot t_i,
\quad t_i=\tfrac{i}4,\quad i=1,2,3,
\end{align}
\begin{align}
\vec{X_iY_i}
&=
Y_i-X_i=
(B-A)\cdot(1-t_i)+(C-D)\cdot t_i
\\
&=
\vec{BA}\cdot(1-t_i)+\vec{CD}\cdot t_i
\\
&=
-\vec{AB}\cdot(1-t_i)-\vec{DC}\cdot t_i
\\
\vec{X_iY_i}
&=
\mathbf{a}\cdot(t_i-1)-\mathbf{b}\cdot t_i
\quad t_i=\tfrac{i}4,\quad i=1,2,3.
\end{align}
