AoPS - Problem

Source: Switzerland - 2020 Swiss MO p7

Tags: geometry, trapezoid, angle bisector, cyclic quadrilateral

parmenides51

08.12.2020 16:07

Let $ABCD$ be an isosceles trapezoid with bases $AD> BC$. Let $X$ be the intersection of the bisectors of $\angle BAC$ and $BC$. Let $E$ be the intersection of$ DB$ with the parallel to the bisector of $\angle CBD$ through $X$ and let $F$ be the intersection of $DC$ with the parallel to the bisector of $\angle DCB$ through $X$. Show that quadrilateral $AEFD$ is cyclic.

Really nice problem. Using the properties of parallel lines, we have $\measuredangle BEX=\frac{\measuredangle DBC}{2}=\measuredangle EXB$, thus $EB=BX$, similarly we have $XC=CF$. By the angle bisector theorem, we have $$\frac{AB}{AC}=\frac{BX}{XC}=\frac{EB}{CF}$$and since $\measuredangle EBA=\measuredangle DBA=\measuredangle DCA=\measuredangle FCA$, we have that $\triangle EBA\stackrel{+}{\sim}\triangle FCA$, hence we have a spiral similarity centred at $A$ taking $EB$ to $FC$, therefore we also have a spiral similarity at $A$ taking $EF$ to $BC$ and since $D=EB\cap FC$, hence we have $(AEF)$ and $(ABC)$ intersecting at $D$. Thus, indeed we conclude that $AEFD$ is cyclic quadrilateral.