Let $\omega$ be a circle. Points $A,B,C,X,D,Y$ lie on $\omega$ in this order such that $BD$ is its diameter and $DX=DY=DP$ , where $P$ is the intersection of $AC$ and $BD$. Denote by $E,F$ the intersections of line $XP$ with lines $AB,BC$, respectively. Prove that points $B,E,F,Y$ lie on a single circle.
Problem
Source: Czech-Polish-Slovak Match 2019 P1
Tags: geometry
15.07.2019 21:10
Since $BD$ is the diameter and $DX=DY=DP$ we have $\sphericalangle DPY = \sphericalangle DPX$, so $$ \sphericalangle YAE = \sphericalangle YAB = 180^{\circ} - \sphericalangle YDB = \sphericalangle DYP + \sphericalangle DPY = \sphericalangle DPY +\sphericalangle DPY = \sphericalangle DPX + \sphericalangle DPY = \sphericalangle YPX = 180^{\circ} - \sphericalangle YPE $$thus $YAEP$ is cyclic. Hence $$ \sphericalangle YBF = \sphericalangle YBC = \sphericalangle YAC = \sphericalangle YAP = \sphericalangle YEP = \sphericalangle YEF $$so $BEFY$ is cyclic too.
15.07.2019 22:00
Czech-Polish-Slovak Match 2019 P1 (Restated) wrote: Let $\Delta ABC$ be an isosceles triangle with $AB=AC$ inscribed in $\omega$. Let $D$ be a point on $\omega$, such, $AD$ is the diameter of $\omega$. Take point $E$ on $AD$, such, $DB=DC=DE$. Let a line through $E$ intersect $\omega$ at $F,G$. Let $BE$ $\cap$ $AG$ $=$ $H$ and $AF$ $\cap$ $BE$ $=$ $I$. Prove, $AHCI$ cyclic. Solution: $\angle AGC=180^{\circ}-\angle ABC=90^{\circ}+\frac12 \angle BAC \implies EHGC$ cyclic. Now, $\angle AFC$ $=$ $\angle ABC$ $=$ $\angle HEC$ $\implies$ $IFEC$ cyclic. $\implies$ $\angle AIC$ $=$ $\angle GEC$ $=$ $\angle GHC$ $\implies$ $AHCI$ is cyclic.
30.09.2019 03:13
ryan17 wrote: Since $BD$ is the diameter and $DX=DY=DP$ we have $\sphericalangle DPY = \sphericalangle DPX$, so $$ \sphericalangle YAE = \sphericalangle YAB = 180^{\circ} - \sphericalangle YDB = \sphericalangle DYP + \sphericalangle DPY = \sphericalangle DPY +\sphericalangle DPY = \sphericalangle DPX + \sphericalangle DPY = \sphericalangle YPX = 180^{\circ} - \sphericalangle YPE $$thus $YAEP$ is cyclic. Hence $$ \sphericalangle YBF = \sphericalangle YBC = \sphericalangle YAC = \sphericalangle YAP = \sphericalangle YEP = \sphericalangle YEF $$so $BEFY$ is cyclic too. Very nice angle chasing!
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