Parabola: Another Interesting Property Let ABCD be a quadrilateral. Suppose there exists a parabola WA with focus A, tangent to lines BC, CD, and DB, and a parabola WC with focus C tangent to lines AB, BD, and AD.
Suppose that WAand WC are tangent to line BD at X and Y respectively. Prove that BX = DY.

I wish to solve the above problem using coordinate geometry
For the sake of simplicity we can assume either one parabola to be y² = 4ax, right? 
Could someone tell me how to proceed, and possibly post a detailed solution for the same? 
I think proving BX = DY can possibly be done using pure geometry/congruence, but, as a student of analytic geometry I wish to proceed using the cartesian coordinate system.
Thanks a lot!  

An analytic proof would necessarily involve parabolas whose axis is not parallel to a coordinate axis, so I think a more geometric approach is simpler.
Let $E$, $F$ be the tangency points of parabola $W_A$ and $G$, $H$ the tangency points of parabola $W_C$ (see diagram below).
It is a well-known property of parabola tangents that
$\angle AEB=\angle ABX$ and $\angle ABE=\angle AXB$.
But $\angle ABE=\angle CBH=\angle CYB$ and $\angle CBY=\angle BAX$: it follows that triangles $ABX$ and $CBY$ are similar and
$$
BX:CY=AX:BY.
$$
By an analogous argument one can show that triangles $ADX$ and $CDY$ are similar and
$$
DX:CY=AX:DY.
$$
Combining the above proportions one gets
$$
BX\cdot BY=DX\cdot DY,
\quad\hbox{that is:}\quad
BX\cdot (BD-DY)=(BD-BX)\cdot DY.
$$
From the last equality, it follows that $BX=DY$.

