Given is triangle $ABC$ with arbitrary point $D$ on $AB$ and central of inscribed circle $I$. The perpendicular bisector of $AB$ intersects $AI$ and $BI$ at $P$ and $Q$, respectively. The circle $(ADP)$ intersects $CA$ at $E$, and the circle $(BDQ)$ intersects $BC$ at $F$ and $(ADP)$ intersects $(BDQ)$ at $K$. Prove that $E, F, K, I$ lie on one circle.
Problem
Source: Serbia JBMO TST 2020 P1
Tags: geometry
leru007
05.09.2020 11:58
$D$ should be an arbitrary point on $ AB$
gnoka
05.09.2020 13:59
It's obvious that C,K,E,F is concylic. So we just need to prove C,I,E,F concylic. Since Triangle AEP equals Tiangle BDP, we have AE=BD. similary we have BF=AD. so CE+CF=AC+BC-AB. Suppose the inscribed circle touches AC, BC at M,N respectively, we have CN=CM=(AC+BC-AB)/2 so EM=FN, which indicates triangle EMI equals triangle FNI, so C,I,E,F concylic. and we are done.
zuss77
05.09.2020 15:46
Probably the same as above. By Miquel $K\in (CEF)$. $AQD\equiv BQF, AEP\equiv BDP$. So $AB=AD+BD=BF+AE$ and by Sparrow lemma $I\in (CEF)$.
StefanSebez
25.09.2021 20:55
$K$ is Miquel point of $\Delta ABC$ so $CEKF$ cyclic it suffices to prove that $CEIF$ is cyclic let $M$ be midpoint of $AB$ let $D'$ be reflection of $D$ with respect to $M$ $\angle AEP=180-\angle PDA=180-\angle PD'D=\angle AD'P$ $\angle PAE=\angle PAD'$ $AP=AP$ $\implies \Delta AEP \cong \Delta APD'$ by $ASA$ $\implies AD'=AE$ so $P$ lies on perpendicular bisector of $ED'$ and so does $I$ $\implies IE=ID'$ similarly $BD'=BF$ and $IF=ID'$ so $IE=ID'=IF \implies I$ is circumcenter of $ \Delta D'EF$ $\angle ECF=\angle ACB=180-\angle EAD'-\angle FBD'=180-(180-2\angle ED'A)-(180-2\angle FD'B)=2(\angle ED'A+\angle FD'B)-180=2(180-\angle ED'F)-180=180-2\angle ED'F=180-\angle EIF$ $\implies CEIF$ is cyclic $\implies EFKI$ is cyclic
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