On the side $BC$ of the parallelogram $ABCD$, points $E$ and $F$ are given ($E$ lies between $B$ and $F$) and the diagonals $AC, BD$ meet at $O$. If it's known that $AE, DF$ are tangent to the circumcircle of $\triangle AOD$, prove that they're tangent to the circumcircle of $\triangle EOF$ as well.
Problem
Source: All-Russian 2021/10.1
Tags: geometry, Angle Chasing
celilcelil
19.04.2021 16:57
With a little angle chasing we can find that CDOF and BAOE are cyclic. It gives us <BDC = <OFE = <DBA = <AEO. Then AE tangents to circumcircle of OEF,similarly DF, then we are done.
rafaello
19.04.2021 20:57
Let $P=AE\cap FD$. Claim. $AOEB$ and $DOCF$ are cyclic. $$\measuredangle BEA=\measuredangle AEF=\measuredangle PEF=\measuredangle PAD=\measuredangle AOD=\measuredangle BOA$$and $$\measuredangle CFD=\measuredangle EFP=\measuredangle ADP=\measuredangle AOD=\measuredangle COD\qquad \square$$ Now, $$\measuredangle OEF=\measuredangle OEB=\measuredangle OAB=\measuredangle CAB=\measuredangle ACD=\measuredangle OCD=\measuredangle OFD,$$thus $PF$ is tangent to $(EOF)$. Similarly, $PE$ is tangent to $(EOF)$. We are done. [asy][asy] size(10cm); import olympiad; import geometry; pair A=dir(100); pair D=dir(200); pair C=dir(0); pair B=A+C-D; dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(45)); dot("$D$", D, dir(D)); pair O=(A+C)/2; dot("$O$", O, dir(O)); draw(A--B--C--D--cycle); path g=circumcircle(A,O,D); pair X=circumcenter(A,O,D); pair F=intersectionpoint(line(B,C),perpendicular(D,line(D,X))); pair E=intersectionpoint(line(B,C),perpendicular(A,line(A,X))); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); path p=circumcircle(E,O,F); draw(g); draw(p); pair P=extension(F,D,E,A); dot("$P$", P, dir(P)); draw(P--E); draw(P--F); draw(F--C); draw(A--C); draw(B--D); draw(circumcircle(D,O,C)); draw(circumcircle(A,O,B)); [/asy][/asy]
hakN
19.04.2021 21:34
$\angle OBE = \angle ADO = \angle OAE \implies AOEB$ is cyclic and similarly $DOFC$ is cyclic. So, $\angle OEF = \angle BAO = \angle BAC = \angle ACD = \angle OCD = \angle OFD \implies AE$ is tangent to $(OEF)$. Similarly we get $DF$ is tangent to $(OEF)$.
L567
19.04.2021 23:06
This one was easy.. $\angle EAO = \angle ODA = \angle EBO$ and so $BEOA$ is cyclic and similarly, so is $CFOD$. Now, to finish, see that $\angle FEO = \angle BAO = \angle OCD = \angle OFD$ and so $FD$ is tangent to $(OFE)$ and similarly, $AE$ is also tangent to it.
cursed_noob
12.09.2021 22:34
Didn't understand this part , $$ \angle EAO = \angle ODA $$Can somebody explain me why the angles are equal??
rafaello
12.09.2021 23:21
cursed_noob wrote: Didn't understand this part , $$ \angle EAO = \angle ODA $$Can somebody explain me why the angles are equal?? This is the Tangent Chord Theorem. In this problem, we know that $AE$ is tangent to $(AOD)$, thus $\angle OAE=\angle ODA$.
WolfusA
12.09.2021 23:22
cursed_noob wrote: Didn't understand this part , $$ \angle EAO = \angle ODA $$Can somebody explain me why the angles are equal?? Kamran011 wrote: If it's known that $AE, DF$ are tangent to the circumcircle of $\triangle AOD$
JAnatolGT_00
04.10.2021 23:24
$$\measuredangle OCF=\measuredangle OAB=\measuredangle OBF, \measuredangle OBE=\measuredangle OAE\implies \measuredangle OEF=\measuredangle OAB=\measuredangle OCD=\measuredangle OFD,\measuredangle OFE=\measuredangle OEA\blacksquare$$
BVKRB-
13.11.2021 15:59
Solved it long back but found the source now
Notice because of the tangency we get $$ \measuredangle OAE = \measuredangle ODA = \measuredangle OBF = \measuredangle OBE \implies \odot(OABE) \ \text{and similarly} \ \odot(ODFC)$$Now $$\measuredangle OAE = \measuredangle OBE = \measuredangle DBE = \measuredangle BDC = \measuredangle ODC = \measuredangle OFE \ \blacksquare $$
REYNA_MAIN
22.12.2021 20:42
$\measuredangle EBO = \measuredangle CBD = \measuredangle ADB = \measuredangle ADO = \measuredangle EAO \implies AOEB$ is cyclic.
$\measuredangle FCO = \measuredangle BCA = \measuredangle DAC = \measuredangle DAO = \measuredangle FDO \implies FODC$ is cyclic.
$\measuredangle AEO = \measuredangle ABO = \measuredangle ABD = \measuredangle CDB = \measuredangle CDO = \measuredangle CFO = \measuredangle EFO \implies AE$ is tangent to $\odot EOF$.
$\measuredangle DFO = \measuredangle DCO = \measuredangle DCA = \measuredangle BAC = \measuredangle BAO = \measuredangle BEO = \measuredangle FEO \implies DF$ is also tangent to $\odot EOF$ and we are done .
Let $AE \cap DF = H$ and FTSOC let $HE$ and $HF$ intersect $\odot EOF$ at $E'$ and $F'$ respectively.
Since $AD \parallel BC = EF$, there exists a homothety centered at $H$ that maps $\odot AOD$ to $\odot EOF$, we also notice that the tangent from $H$ intersects $\odot AOD$ at $\{A, A\}$ and $\{D, D\}$ (considering $AA$ w.r.t. $\odot AOD$ means the tangent at $A$ to $\odot AOD$) respectively. Thus $A$ gets mapped to $E$, $D$ to $F$, $A$ to $E'$ and $D$ to $F' \implies E' = E$ and $F' = F \stackrel{CONTRADICTION}{\implies} HE$ and $HF$ are tangent to $\odot EOF$ and we are done maybe?
PLEASE SOMEONE CHECK MY SECOND SOLUTION, ITS A NEW ONE, AND I'M NOT SURE IF ITS A CORRECT ONE. PLEASE GO THROUGH IT ONCE IF U HAVE TIME .
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