$E,F$ are points on $AB,AC$ that satisfies $(B,E,F,C)$ cyclic. $D$ is the intersection of $BC$ and the perpendicular bisecter of $EF$, and $B',C'$ are the reflections of $B,C$ on $AD$. $X$ is a point on the circumcircle of $\triangle{BEC'}$ that $AB$ is perpendicular to $BX$,and $Y$ is a point on the circumcircle of $\triangle{CFB'}$ that $AC$ is perpendicular to $CY$. Show that $DX=DY$.
Problem
Source: 2021 Korea Winter Program Practice Test
Tags: geometry, perpendicular bisector, circumcircle
KPBY0507
09.02.2021 09:30
Akatsuki1010 wrote: Something is wrong here. $B',C'$ are the reflections, not the foot of the perpendicular line.... I think $B',C'$ are the foot of the perpendicular line to $AD$ from $B,C$ in your picture.
L.Bill
11.02.2021 02:37
Something is wrong here, I draw the figure and find that $DX\neq DY$.
L.Bill
11.02.2021 02:55
Maybe is $AB\perp BX, AC\perp CY$, then the problem is right.
space10
14.02.2021 19:34
Let $l$ a radical axis of $\odot (BEC'), \odot (CFB')$. $AE\cdot AB=AF\cdot AC\implies A\in l, BD\cdot CD=B'D\cdot C'D\implies D\in l, AD=l$ $BC', B'C$ are symmetrical to $AD\implies \odot (BEC'), \odot (CFB')$are symmetrical to $AD$. Let $O_1$ a center of $\odot (BEC')$, $O_2$ a center of $\odot (CFB')$. $O_1, O_2$ are symmetrical to $AD$ so we get $DO_1=DO_2$. $EO_1=radius(\odot (BEC'))=radius(\odot (BEC'))=FO_2$. By problem, $ED=FD \implies \triangle{EDO_1} \equiv \triangle{FDO_2}$ (SSS) $\angle{DEX} =\angle{DEO_1} =\angle{DFO_2}= \angle{DFY}$ $EX, FY$ is diameter of $\odot (BEC'), \odot (CFB')\implies EX=FY$ $\therefore \triangle{DEX} \equiv \triangle{DFY}$(SAS), so we get $DX=DY$. Q.E.D.
space10
14.02.2021 19:36
L.Bill wrote: Maybe is $AB\perp BX, AC\perp CY$, then the problem is right. I agree with that.
Ali3085
06.03.2021 23:25
let $E',F',X',Y'$ be the reflection of $E,F,X,Y$ wrt $AD$ note that $AE'.A'B=AE.AB=AF.AC$ so $FE'B'C$ is cyclic so $XX'YY'$ is an isoscele trapezium claim: $D$ is the center of $(XX'YY')$ proof: since $\angle X'B'A=90$ wee have $E'FX'Y$ is a rectangle since $DE=DE'=DF$ so $DX'=DY$ similarly $DY'=DX$ and we win
gnoka
08.04.2021 06:18
space10 wrote: $BD\cdot CD=B'D\cdot C'D\implies D\in l, AD=l$ but $BD\cdot CD$ is not the power of D to $\odot (BEC') or \odot (CFB')$, so this does not make sense.