I am looking for a previous reference and/or a geometric proof of the following lemma:
Let $P$ be the parabola $y=x^2$. Let $a$, $b$, $c$, $d$ be four points on $P$ sorted from left to right, and let $z$ be the point of intersection of the segments $ac$ and $bd$. Define the horizontal distances $p=b_x-a_x$, $q=d_x-c_x$, $r=z_x-b_x$, $s=c_x-z_x$. Then, $p/q=r/s$.
This is easily proven algebraically, but I was wondering whether there is a geometric argument.
It is equivalent to prove $\frac {r}{p} = \frac {s}{q}$ .
In the green shaded region, by intercept theorem, we have $\frac { ZC’}{C’D} = \frac {s}{q}$.
Similarly, in the yellow region, we have $\frac {ZB’}{B’A} = \frac {r}{p}$ .
It remains to show is $B’C’$ parallel $AD$. [A geometric interpretation of the result]
The above must be true because the equality of the said ratios has been proven algebraically.