# 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!

• This is very nice property! Similar to incircle and excircle. I wonder if this is somehow valid also for ellipse and hyperbola. – Maria Mazur Dec 28 '17 at 12:59
• Yeah, I know right! I wonder so too! – arya_stark Dec 28 '17 at 12:59
• Is $ABCD$ cyclic? It seem that it is necessary for such configuration. – Maria Mazur Dec 28 '17 at 14:24
• @John Watson: Yes, the quadrilater ABCD must to be cyclic: this can proved by applying Desargues involution theorem. – Fabio Lucchini Dec 28 '17 at 23:23
• @Fabio Lucchini Actualy we don't need involution here, it is simple consequence of Simson's theorem. If we have three tangents with intersections A,B,C, then the focus of parabola is on circumcircle of triangle $ABC$. cut-the-knot.org/Curriculum/Geometry/… – Maria Mazur Dec 29 '17 at 10:00

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 $BX=DY$.
• BTW, the above shows, as an easy corollary, the result mentioned in the comments to the question: $\square ABCD$ is cyclic. (Performing the same angle analysis on the right-hand side of the diagram, one sees that the opposite angles of the quadrilateral must be supplementary.) – Blue Dec 30 '17 at 9:42