In a convex quadrilateral $ABCD$, the lines $AB$ and $DC$ intersect at point $P{}$ and the lines $AD$ and $BC$ intersect at point $Q{}$. The points $E{}$ and $F{}$ are inside the quadrilateral $ABCD$ such that the circles $(ABE), (CDE), (BCF),(ADF)$ intersect at one point $K{}$. Prove that the circles $(PKF)$ and $(QKE)$ intersect a second time on the line $PQ$.
Problem
Source: Russian TST 2020, Day 6 P3
Tags: geometry
PROF65
24.03.2023 18:39
Lemma If $ABCD$ is a quadrilateral and $X,Y,Z,W$ are points on $AB,BC,CD,DA$ respectively s.t. $(WAX),(XBC),(YCZ),(ZDW)$are concurent at $M$ then $1) $ if $\{M,E\}=(WAX)\cap(YCZ), \{M,F\}=(XBY)\cap(ZDW)$ then $A,E,D$ and $B,F,C$ are collinear. $2)$ $XY,ZW,AC$ are concurent $ \iff XW,BD,YZ$ are concurent $3)$ if $AC\cap BD=G$ then $EFGM$ is cyclic . proof $1)$ just using Miquel point with $ABD$ and $BCA$. $2)$ using Menalaus's we get that the concurency of $XY,ZW,AC$ and of $XW,BD,YZ$ means $\frac{XB}{XA}.\frac{WA}{WD}.\frac{ZD}{ZC}.\frac{YC}{WB}=1$ $3)$$\angle (GE,GF)=\angle (AE,AD)+\angle (AD,DF)=\angle (ME,MW)+\angle (MW,MF)=\angle (ME,MF)$ hence $MGEF$ is cyclic . Back to the Problem Let $(KPF)\cap (QKE)=\{K,H\}$ , $PF$ cuts $QE$ at $G$ we'll show that it equivalent to prove $KEFG$ is cyclic indeed : $\angle (PH,HK)=\angle (PF,KF)=\angle (FG,FK),\angle (QH,HK)=\angle (QE,KE)=\angle (EG,EK)$ which concludes the claim. Now Let $PF$ cuts again $(BFC),(ADE)$ at $R,S $ ; $RC$ hits $ SD$ at $ U$ ; $RB$ hits $ SA$ at $ V$ we have $\angle (KA,AV)=\angle (KA,AS)=\angle (KF,FS), \angle (KB,BV)=\angle (KF,FR) $ hence $ABKV $ and also $ABKVE $ are cyclic similarly we get $DCKUE $ is cyclic so applying the lemma with $SURV$ and the points $D,C,B,A$ we deduce $KEFG$ is cyclic which ends the proof . Best regards RH HAS
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