What is the ratio of $\frac{XY}{SQ}$ In the picture,$PQRS$ is a prallelogram. $PS$ is parallel to $ZX$ and $\frac{PZ}{ZQ}=\frac{2}{3}$. Then$\frac{XY}{SQ}$ equals:

The answer is $\frac{9}{40}$
Help me with the idea to solve this problem.
Thanks in advance.
 A: I will outline the proof and you may add back some details (i.e. reasons) to it.
$ΔZYQ \sim ΔSYR$ $⇒ ZY/YR=3/5$ and $YQ/SY=3/5$
$ΔXYZ \sim ΔYQR$ $⇒ XY/YQ=3/5$ (transitivity)
$XY/SY=(YQ/SY)/(YQ/XY)=9/25$
$YQ/XY=15/9$ and $XY/SY=9/25$
$∴ XY/SQ=9/(SY+YQ)=9/(25+15)=9/40$ 
A: This can be brute-forced using analytical geometry. 
We start with 
$$
\frac{2}{3}
= \frac{\lVert PZ\rVert}{\lVert ZQ\rVert} 
= \frac{\lVert SX\rVert}{\lVert XQ\rVert}
$$
We want the ratio
$$
\frac{\lVert XY\rVert}{\lVert SQ\rVert}
$$
and use the linear interpolation formula
$$
C = (1-t) A + t B \quad (t \in [0,1])
$$
for a point $C$ on a line between points $A$ and $B$ to observe
$$
X = (1-3/5) Q + 3/5 S = 2/5 Q + 3/5 S \\
Y = (1-\lambda) Q  + \lambda S = (1-\mu) Z + \mu R
$$
If we knew the coordinates for all points this should be solvable. For this  we choose lengths for a particular instance of all scalings, choose $Q$ as origin (so it drops from the above formulas) and have positive coordinates to the left  (to not worry about negative values too much):
$$
Q = (0,0), \quad P = (5, 0), \quad Z = (3, 0)
$$
This gives
$$
X = (x, y), \quad S = ((5/3)x, (5/3)y)
$$
For a parallelogram we have
$$
R = S + PQ = ((5/3)x, (5/3)y) + (-5, 0) = (5/3)x - 5, (5/3) y)  
$$
The equation for $Y$ reduces to:
$$
Y = \lambda S = \lambda ((5/3)x, (5/3)y) = (1 - \mu) (3,0) + \mu ((5/3)x-5,(5/3)y)
$$
which gives for the components:
$$
(5/3)x\lambda = 3 - 3\mu + (5/3) x \mu - 5\mu, \quad 
(5/3)y\lambda = (5/3) y \mu
$$
thus 
$$
\lambda = \mu, \quad 
(5/3)x\lambda = 3 - 3\lambda + (5/3) x\lambda -5 \lambda \iff 
\lambda = 3/8
$$
and
$$
Y = \lambda S = ((5/8)x,(5/8)y)
$$
which gives
$$
XY = X - Y = ((3/8)x, (3/8)y) \\
SQ = S - Q = ((5/3)x, (5/3)y) 
$$
So
$$
\lVert XY\rVert^2 = (3/8)^2 (x^2 + y^2) \\
\lVert SQ \rVert^2 = ((5/3)x, (5/3)y) = (5/3)^2 (x^2 + y^2)  
$$
and 
$$
\frac{\lVert XY\rVert}{\lVert SQ\rVert} = \frac{3/8}{5/3} = \frac{9}{40}
$$
A: Hint : $\Delta XYZ ,\Delta YQR$ are similar 
$\Delta YZQ , \Delta YRS$ are similar. 
Gives 
$\frac{XY}{YQ}=\frac{YZ}{YR}$
and $\frac{YZ}{YR}=\frac{ZQ}{RS}=\frac{3}{5}$
Further Gives us $\frac{YQ}{XY}=\frac{5}{3}$
or $\frac{YQ+XY}{XY}=\frac{8}{3}$
I am sure you can see where this is going from now!
