A parallelogram ABCD circumscribes an ellipse at P,Q,R and S. Prove $\frac{CQ}{QB}=\frac{CR}{BP}$ I found this statement in my notes with only a figure as proof .
We have ABCD a parallelogram , E is an ellipse 
E is tangent to each side AB , BC , CD , DA to the points P,Q,R,S. So we have:
$$\frac{CQ}{QB}=\frac{CR}{BP}$$
I thought that maybe using the equation for the ellipse & tangent may help but it doesn't lead me anywhere 
Thanks in advance with your help with this proof 
 A: Apply a linear transformation that turns the ellipse into a circle; let $X'$ be the image of $X$ under the transformation ($A$ becomes $A'$, etc). Linear transformations preserve parallelism, and they preserve ratios of lengths of parallel segments, so $$\frac{|CQ|}{|QB|}=\frac{|C'Q'|}{|Q'B'|} \qquad \frac{|CR|}{|BP|}=\frac{|C'R'|}{|B'P'|}$$ But
$|B'P'|=|B'Q'|$ and $|C'Q'|=|C'R'|$, as lengths of congruent tangent segments from $B'$ and $C'$ to the circle $E'$. The result follows. $\square$
A: 
Apply the sine rule to the triangles BPQ and CRQ,
$$\frac{BP}{BQ}=\frac{\sin\alpha}{\sin\gamma}=\frac{\sin(\theta+\gamma)}{\sin\gamma}
=\sin\theta(\cot\gamma+\cot\theta)\tag{1}$$
$$\frac{CR}{CQ}=\frac{\sin\beta}{\sin\delta}=\frac{\sin(\theta-\delta)}{\sin\delta}
=\sin\theta(\cot\delta-\cot\theta)\tag{2}$$
From the ellipse equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we can get the tangent of the line BC,
$$\cot\theta = \frac{1}{y’}=-\frac{b^2y}{a^2x}$$
Also, the slopes of the line PQ and RQ,
$$\cot\gamma=-\frac{x}{y-b}, \>\>\>\>\>\cot\delta=\frac{x}{y+b}$$
Now, evaluate
$$\cot\gamma+\cot\theta = -\frac{x}{y-b} -\frac{b^2y}{a^2x}= \frac{a^2}{bx}$$
$$\cot\delta-\cot\theta = \frac{x}{y+b} +\frac{b^2y}{a^2x}= \frac{a^2}{bx}$$
which have the same value. Plug above results into (1) and (2) to obtain,
$$\frac{BP}{BQ}=\frac{CR}{CQ}$$
