Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$ lying in the interior. Let $E$ and $F$ be the midpoints of the segments $BC$ and $AD$, respectively. Let $X$ be the point lying on the same side of the line $EF$ as the vertex $C$ such that $\triangle EXF$ and $\triangle BOA$ are similar. Prove that $XC = XD$.
Problem
Source: BMO SL 2023 G2
Tags: geometry, BMO Shortlist
VicKmath7
03.05.2024 15:31
Let $Q$ be the Miquel point of $ABCD$ and let $AB \cap CD=P$. If $M$ is the midpoint of $CD$, then we are asked to show that $O, X, M$ are collinear. Since $\angle OMP=\angle OQP=90^{\circ}$, we obtain that $OMQP$ is cyclic and $\angle MOQ=\angle MPQ=\angle DAQ$, so we are left to show that $\angle QOX=\angle QAD$. By spiral similarity at $Q$, it's easy to see that $\triangle QEF \cup \{X\} \sim \triangle QBA \cup \{O\}$, i.e. $\triangle QOX \sim \triangle QAF$, which implies the desired angle equality.
Assassino9931
03.05.2024 16:44
Good looking problem, but too bad it is trivial by complex bash.
Om245
03.05.2024 22:11
Let $Q$ be Miqual point of $\Box ABCD$ with $X= AB \cap CD$. Let $M$ be midpoint of $CD$. Note by similarity condition we have $AO \rightarrow FE$. Define $Y= FA \cap OX$. Then by property we have $Q,Y,A,O$ cyclic $$\measuredangle AYO = \measuredangle AQO = 90 - \measuredangle XQA = 90 - \measuredangle CDA$$hence $OX \perp CD$ , but we know $OM \perp CD$. Hence $M$ lie on $OX$ and $XC=XS$.
Circumcircle
03.05.2024 22:38
This problem was proposed by me. I know there are solutions by complex numbers and by Mean Geo etc, but I am posting here mine. Let $M$ be the midpoint of $CD$. Let $EM$ intersect $AD$ at $Z$ and $FM$ intersect $BC$ at $Y$. As $F$, $M$, and $E$ are midpoints of $AD$, $DC$, and $BC$ respectively, we have that $FM||AC$ and $EM||BD$. $\angle EZF=\angle BDA=\angle ACB=\angle FYE\Rightarrow FEYZ$ is cyclic. Note that since $\bigtriangleup EXF\sim \bigtriangleup BOA$, we have that $XF=XE$ and also that $\angle EXF=\angle BOA=2\angle ACB=2\angle FYE\Rightarrow X$ is the center of $FEYZ$. Since $MD=MC$, apply Converse of Butterfly Theorem and get that $XM\perp DC\Rightarrow XD=XC$.
matematica007
04.05.2024 15:54
Trivial solution by complex bash. Let circumcircle of $ABCD$ be the unit circle on the complex plane and let $A(a)$,$B(b)$,$C(c)$,$D(d)$ ,$X(x)$ be the coordinates of the points in the complex plane. We have that $F(\frac{a+d}{2})$ and $E(\frac{b+c}{2})$. How $\triangle EXF$ and $\triangle BOA$ are similar we have $\frac{a}{\frac{a+d}{2}-x}=\frac{b}{\frac{b+c}{2}-x}$ so $ac-2xa=bd-2xb$ and this implies that $x=\frac{ac-bd}{2a-2b}$ so $\overline{x}=\frac{bd-ca}{cd \cdot(2b-2a)}$ and this implies that $x=cd\cdot \overline{x}$. $XC = XD$ is equivalent to $|x-d|=|x-c|$ and this is equivalent to $x=cd\cdot \overline{x}$. So the problem is proved.
Z4ADies
24.07.2024 14:31
From similarity $XO,FA,EB$ concurr at point $Y$. $\angle AYE=\angle OEF$ $\implies$ $\angle FEY=90-\angle AYE=\angle FDC$.Thus,$YX $ is perpendicular to $DC$. Also $OD,OC$ that means $XO$ is perpendicular bisector of $DC$.Which is clear that $XD=XC$.
Nari_Tom
30.01.2025 06:57
Interesting solution using Gliding-Principle. Let $O'$ be the reflection of $O$ w.r.t point $X$. Then By the gliding principle we have: $\triangle BAO \sim \triangle EFX \sim \triangle CDO'$. Which implies that $O'$ lies on the perp bisector of $CD$ $\implies$ $X$ lies on it.