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 $BX=DY$.