Let $ABCD$ be a convex quadrilateral with $\angle ABC=\angle ADC=120^{\circ}$. The point $E$ lies on the segment $AD$ and is such that $AE \cdot BC=AB \cdot DE$ and similarly the point $F$ lies on the segment $BC$ and satisfies $BF \cdot CD=AD \cdot CF$. Show that $BE$ and $DF$ are parallel.
Problem
Source: Polish MO Second round 2024 P2
Tags: geometry
math_comb01
09.02.2024 23:23
Sketch : Construct equilateral triangle $BCC'$, then re-define $F = DC' \cap BC$ and length chase.
aqwxderf
09.02.2024 23:29
Let the bisector of $\measuredangle ABC$ intersect $\overline{AC } $ at $X $, and let the bisector of $\measuredangle ADC$ intersect $\overline{ AC} $ at $Y $. By the angle bisector theorem, we have: $XA : XC = BA : BC = EA : ED $ thus we have $EX \parallel DC$. Similarly, we have $FY \parallel BA$. $\measuredangle ABX = 60^\circ = \measuredangle ADC = \measuredangle AEX$, thus $ABXE $ is cyclic. Similarly, $CDYF $ is cyclic. $\measuredangle DFC = \measuredangle DYC = \measuredangle DAC + \measuredangle ADY = \measuredangle EBX + 60^\circ = \measuredangle ADY + \measuredangle XBC = \measuredangle EBC $ thus $BE \parallel DF$.
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