I had done a same similar problem a few days back Anyways, here is a solution: $\angle EDF = 180^{\circ} - \angle EDB -\angle FDC = 180^{\circ} - 2A = 180^{\circ} - \angle EOF$. Thus $OEDF$ is cyclic. But we had $OE = OF$, so by Fact 5/incenter - excenter lemma, $OD$ bisects $\angle EDF$.

TheDarkPrince wrote: I had done a same similar problem a few days back Anyways, here is a solution: $\angle EDF = 180^{\circ} - \angle EDB -\angle FDC = 180^{\circ} - 2A = 180^{\circ} - \angle EOF$. Thus $OEDF$ is cyclic. But we had $OE = OF$, so by Fact 5, $OD$ bisects $\angle EDF$. What's fact 5, btw and why is it okay to just cite it?

http://web.evanchen.cc/handouts/Fact5/Fact5.pdf You should probably call it the "incenter-excenter lemma" when you cite it.

Why is $\measuredangle EDB$ equal to $\measuredangle A$?

@above Since, Points $A,E,D,C$ are concyclic

Couldn’t find a better solution; But what does RMO stand for? Rioplatense MO?

Regional Mathematics Olympiad

Answering one year old question Note that $BDFA$ and $CDEA$ are cyclic quadrilaterals. Thus by power of point, we can conclude that $CF\cdot CA=CD\cdot CA$ and $BE\cdot BA=BD\cdot BC$. Clearly, $CF\cdot CA$ is the power of $C$ w.r.t. $(AEF)$ and thus, $$CF\cdot CA=CO^2-r^2=CD\cdot CB=CD^2+CD\cdot DB$$Similarly,$$BO^2-r^2=BD^2+BD\cdot DC$$Hence, $CO^2-CD^2=BO^2-BD^2\implies OD\perp BC$. Also, $\angle BAC=\angle EDB=\angle FDC$ and thus $\angle EDO=\angle FDO$.

We have: $\angle{EDF} = 180^o - \angle{BDE} - \angle{CDF} = 180^o - 2 \angle{BAC} = 180^o - \angle{EOF}$ Then: $O$, $E$, $D$, $F$ lie on a circle But: $OE = OF$ so: $DO$ is bisector of $\angle{EDF}$

khanhnx wrote: We have: $\angle{EDF} = 180^o - \angle{BDE} - \angle{CDF} = 180^o - 2 \angle{BAC} = 180^o - \angle{EOF}$ Then: $O$, $E$, $D$, $F$ lie on a circle But: $OE = OF$ so: $DO$ is bisector of $\angle{EDF}$ Why $\angle{BDE} + \angle{CDF} = 2\angle{BAC}$ ?

Claim: $OEDF$ is cyclic. Proof: By Inscribed Angle theorem, $\angle EOF = 2 \angle BAC$. Now, since $ABDF$ is cyclic, $\angle BDF = 180 - \angle BAC$ and similarly since $ACDE$ is cyclic, $\angle CDE = 180 - \angle BAC$, but $\angle CDE + \angle BDF - \angle EDF = 180 \implies \angle EDF = 180 - 2 \angle BAC = 180 - \angle EOF \implies OEDF$ is cyclic. But now $OE = OF$ so $\angle OEF = \angle OFE = \angle EDO = \angle FDO \implies OD$ bisects $\angle EDF$

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TheDarkPrince wrote: Let $ABC$ be an acute angled triangle and let $D$ be an interior point of the segment $BC$. Let the circumcircle of $ACD$ intersect $AB$ at $E$ ($E$ between $A$ and $B$) and let circumcircle of $ABD$ intersect $AC$ at $F$ ($F$ between $A$ and $C$). Let $O$ be the circumcenter of $AEF$. Prove that $OD$ bisects $\angle EDF$.

Can it solve with inversion

Clean Angle Chase (recognized this from an OMO problem ) $\angle ABC = \angle DFC = \angle B$ (Exterior angle of a cyclic quadrilateral) Similarly $\angle ACB = \angle DEB = \angle C$ $\implies \angle EDC = \angle FDC = \angle A$ $\implies \angle EDF = 180^\circ - 2\angle A$ But notice that $\angle EOF = 2\angle A$ (Angle subtended at the center) $\implies \angle EOF + \angle EDF = 180^\circ$ Hence Quadrilateral $EDFO$ is cyclic Also note that $\angle OEF = \angle OFE = x$ ($OE = OF$) $\angle ODF = \angle OEF = x$ (Angle subtended by same segment) $\angle ODE = \angle OFE = x$ (Angle subtended by same segment) $\implies \angle EDO = \angle FDO = x$ $\implies OD$ bisects $\angle EDF \ \blacksquare$

My solution is same as khan hnx

The geo was easy,or May be it looked easy after I spotted the trick

so first of since points $A,F,D,B$ are concyclic we get $\angle{DFC}=B$ and similarly we have points $A,E,D,C$ to be concyclic so we get $\angle{BED}=C$ so we notice that $\angle{EDB}=\angle{FDC}=A$ also in quadrilateral $AEDF$ we get $\angle{EDF}=180-2A$ and if we join $OE$ and $OF$ we get $\angle{EOF}=2A$ which gives $O,E,D,F$ to be cyclic and hence $\angle{FEO}=\angle{FDO}$ similarly $\angle{OFE}=\angle{ODE}$. but since $\angle{FEO}=\angle{OFE}$ we get $\angle{FDO}=\angle{ODE}$ which gives that $O$ bisects $\angle{EDF}$ $\blacksquare$

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Claim: Points $E,O,F,D$ are concyclic. Proof: $\angle AFD=180^{\circ}-\angle B$ and $\angle AED=180^{\circ}-\angle C$. Now using the interior angle sum property for quadrilateral $AEDF$, we get that $\angle EDF=\angle B+\angle C-\angle A$. Also, $\angle EOF=2\angle EAF=2\angle A$. Thus, $\angle EOF+\angle EDF=\angle A+\angle B+\angle C=180^{\circ}$, which completes our proof. Now $\angle OEF=\angle OFE$. But $\angle OEF=\angle ODF$ and $\angle OFE=\angle ODE$. Thus $\angle ODE=\angle ODF$. Our proof is thus complete.