A parallelogram ABCD circumscribes an ellipse at P,Q,R and S. Prove $\frac{CQ}{QB}=\frac{CR}{BP}$

Question

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

Answer 1

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. $\square$

Answer 2

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,

$$\frac{BP}{BQ}=\frac{CR}{CQ}$$