Chord $D E$ of the circumcircle of the triangle $A B C$ intersects sides $A B$ and $B C$ in points $P$ and $Q$ respectively. Point $P$ lies between $D$ and $Q$. Angle bisectors $D F$ and $E G$ are drawn in triangles $A D P$ and $Q E C$. It turned out that the points $D$, $F, G, E$ are concyclic. Prove that the points $A, P, Q, C$ are concyclic. Azamat Mardanov
Problem
Source: 45th International Tournament of Towns, Senior A-Level P5, Fall 2023
Tags: geometry
NO_SQUARES
11.12.2023 14:32
Nice problem!
In fact, if $D, F, G, E$ are concyclic, then $B$ is center of this circle.
SBYT
12.12.2023 09:24
$\angle DEG=\frac{1}{2}\angle DEC=\frac{1}{2}\angle DBC$and$\angle EDF=\frac{1}{2}\angle EDA=\frac{1}{2}\angle EBA$. So $B$ is the center of $\odot DFGE$,especially we have $BD=BE$.$\angle EQC=\frac{1}{2}(\stackrel\frown{EC}+\stackrel\frown{BD})=\frac{1}{2}(\stackrel\frown{EC}+\stackrel\frown{BE})=\stackrel\frown{BC}=\angle BAC$,so $A,P,Q,C$ are concyclic.
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NO_SQUARES
12.12.2023 15:52
SBYT wrote: $\angle DEG=\frac{1}{2}\angle DEC=\frac{1}{2}\angle DBC$and$\angle EDF=\frac{1}{2}\angle EDA=\frac{1}{2}\angle EBA$. So $B$ is the center of $\odot DFGE$ Can you say more about it, please? Why $B$ can't be second intersection $(GOD)$ and $(FOE)$, there $O$ is center of $(DEGF)$?
gnoka
14.12.2023 09:50
SBYT wrote: $\angle DEG=\frac{1}{2}\angle DEC=\frac{1}{2}\angle DBC$and$\angle EDF=\frac{1}{2}\angle EDA=\frac{1}{2}\angle EBA$. So $B$ is the center of $\odot DFGE$,especially we have $BD=BE$.$\angle EQC=\frac{1}{2}(\stackrel\frown{EC}+\stackrel\frown{BD})=\frac{1}{2}(\stackrel\frown{EC}+\stackrel\frown{BE})=\stackrel\frown{BC}=\angle BAC$,so $A,P,Q,C$ are concyclic. 我去……你们能不能不要id整活的同时头像暴露真实身份