Let $\Gamma$ be the circumcenter of $\triangle ABC$. Then I claim that $O = \Gamma \cap AI$ (intersection that's not $A$ obviously) Let $F = \Gamma \cap AI$. As $F$ lies on the angle bisector of $\angle BAC$, it follows $F$ is the midpoint of the minor arc $BC$ so $BF = FC$. Now we seek to show $FB=FI$. Note that $\angle BFI = \angle BFA = \angle BCA$. However, $\angle FBC = \angle FCB = \frac{180 - \angle BFC}{2} = \frac{\angle BAC}{2}$ Notice $\angle IBF = \frac{\angle BAC + \angle ABC}{2}$ We quickly derive then that $\angle BIF = \frac{\angle BAC + \angle ABC}{2} = \angle IBF$. Thus $FB=FI=DC$, hence $F$ is the circumcenter of $\triangle BIC$ so it follows $F=O$ lies on $AI$. But then $\angle DAO = \angle EAO$ and then since $AD = AE$, we have $\triangle ADO \cong \triangle AEO \implies \angle ADO = \angle AEO \implies \angle OBD = \angle OEC$ as desired.

Let $\angle {ODB}=\theta,\angle {OEC}=\alpha.$ So $\frac {OE}{OD}=\frac {Cos (A/2+\theta)}{Cos (B/2+\alpha)}$ Also $\frac {OE}{OD}=\frac{Cos(\theta)}{Cos(\alpha)}$ ,Combining done.

Clearly, $I, O$ belong to angle bisector of $\angle BAC$, while $D,E$ are symmetrical about it, done. Best regards, sunken rock

Let $AI$ intersect $\odot (ABC)$ again at $O'$. Since $AI$ is the internal bisector of $\angle A,O'B=O'C=O'I \Rightarrow O'$ is the circumcircle of $\triangle BCI \Rightarrow O'=O$. $A,I,O$ are collinear.$AD=AE$ and $\angle IAD=\angle IAE \Rightarrow D,E$ are reflections of each other in $AO$. Hence $\angle ADO=\angle AEO \Rightarrow \angle ODB=\angle OEC$

Let $\frac{A}{2} = \alpha$, $\frac{B}{2} = \beta$ and $\frac{C}{2} = \gamma$. $\angle BIC = \angle ABI + \angle ACI + \angle BAC = 2\alpha + \beta + \gamma = 90^\circ + \alpha$ so $\angle BOC = \angle BOI + \angle COI = 2(\angle ICB + \angle IBC) = 2(180^\circ - \angle BIC) = 180^\circ - 2\alpha$ so $ABOC$ is cyclic. Thus $O$ is the intersection point (closer to $I$ than $A$) of the perpendicular bisector of $BC$ with the circumcircle of $ABC$. Thus $A$, $I$, $O$ are collinear. Then by SAS, $DAO$ and $EAO$ are congruent so $\angle ADO = \angle AEO$ so $\angle ODB = \angle OEC$.

No pencil, no paper. We would first show that proving AIO collinear is sufficient. Since ADIE is a kite, I is on the perpendicular bisector of DE, so O lies on the perpendicular bisector of DE. Then DO=OE, so ADOE is a kite: <ADO=<AEO so <ODB=<OEC. Now we would show that AIO is collinear. <AIB=180-A/2-B/2. <BCI=C/2, so <BOI=C. Since BO=OI, <BIO=90-C/2. Thus <AIB+<BIO=180-A/2-B/2+90-C/2=180. Hence AIO is collinear and we are done.

Well known that $A,I,O$ are collinear. Now $\triangle ADO \cong \triangle AEO \implies \angle ODB = \angle OEC$ as desired.$\blacksquare$

Dear Mathlinkers, also at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32163 Sincerely Jean-Louis

This is trivial if you know Chicken Feet Theorem beforehand

@above what is that?

Question: Can we use the incenter-excenter lemma directly on contests?

Isn't it called symmetry wrt. $AI$?

By the Incenter Excenter Lemma, $A, I$, and $O$ are collinear. Now, we note that $DI=EI$ and $IO=IO$ trivially. Furthermore, also note that $$\angle DIO=180^\circ-\angle AID=180^\circ-(90^\circ-\angle DAI)=90+\angle DAI \ \ \text{and} \ \ \angle EIO=180^\circ-\angle AIE=180^\circ-(90^\circ-\angle EAI)=90^\circ+\angle EAI=90^\circ+\angle DAI.$$This implies that $\triangle DIO\cong\triangle EIO$ which in turn, implies $\angle ODI=\angle OEI$. Finally, $$\angle ODB=90^\circ-\angle ODI \ \ \text{and} \ \ \angle OEC=90^\circ-\angle OEI=90^\circ-\angle ODI$$as desired. $\blacksquare$

Storage. By fact 5, $A,I,O$ are on the $A$ angle-bisector. Note that $AD=AE,\measuredangle DAO=\measuredangle OAE,$ and $AO=AO$. So by SAS Congruence, we are done.

Just Even Chen’s incenter excenter theorem

v_Enhance wrote: As shown in the figure below, the in-circle of $ABC$ is tangent to sides $AB$ and $AC$ at $D$ and $E$ respectively, and $O$ is the circumcenter of $BCI$. Prove that $\angle ODB = \angle OEC$. [asy][asy]import graph; size(5.55cm); pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-5.76,xmax=4.8,ymin=-3.69,ymax=3.71; pen zzttqq=rgb(0.6,0.2,0), wwwwqq=rgb(0.4,0.4,0), qqwuqq=rgb(0,0.39,0); pair A=(-2,2.5), B=(-3,-1.5), C=(2,-1.5), I=(-1.27,-0.15), D=(-2.58,0.18), O=(-0.5,-2.92); D(A--B--C--cycle,zzttqq); D(arc(D,0.25,-104.04,-56.12)--(-2.58,0.18)--cycle,qqwuqq); D(arc((-0.31,0.81),0.25,-92.92,-45)--(-0.31,0.81)--cycle,qqwuqq); D(A--B,zzttqq); D(B--C,zzttqq); D(C--A,zzttqq); D(CR(I,1.35),linewidth(1.2)+dotted+wwwwqq); D(CR(O,2.87),linetype("2 2")+blue); D(D--O); D((-0.31,0.81)--O); D(A); D(B); D(C); D(I); D(D); D((-0.31,0.81)); D(O); MP( "A", A, dir(110)); MP("B", B, dir(140)); D("C", C, dir(20)); D("D", D, dir(150)); D("E", (-0.31, 0.81), dir(60)); D("O", O, dir(290)); D("I", I, dir(100)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Wot? Ez. Claim 1 : $\triangle ADO \cong \triangle AEO$ Due to fact $5$ it is well known that $O$ lies on line $\overline{AI}$, therefore $\angle BAI = \angle CAI \implies \angle DAO = \angle EAO$, segment $AO$ is common to both $\triangle ADO$ and $\triangle ACO$ and finally, $AD = AE$ as line $\overline{AD}$ and line $\overline{AE}$ are both tangent to the incircle $\Gamma$ of $\triangle ABC$, therefore due to $SAS$ test we have that $\triangle ADO \cong \triangle AEO$ We know that corresponding of congruent angles have same measure, therefore $\angle ADO = \angle AEO \implies 180 - \angle BDO = 180 - \angle CEO \dots (1)$ where $(1)$ is a consequence of the fact that $D \in$ line $\overline{AB}$ and $E \in$ line $\overline{AC}$ and so due to $(1)$, we have that $\angle BDO = \angle CEO$ as desired.

From the Incenter-Excenter Lemma, we know that $A,I,O$ are collinear. Notice that $\angle IDB=\angle IEC=90$. Because $\angle IDA=\angle IEA=90$ and $\angle IAD=\angle IAE$, $\angle AID=\angle AIE$, so $\angle DIO=\angle EIO$. Combining $DI=IE$ and $IO=IO$, we have $\triangle DIO\cong \triangle EIO$. Thus, $\angle IDO=\angle IEO$, and $\angle ODB=\angle OEC$ as desired. $\blacksquare$

By incenter-excenter lemma we have that $O$ is the midpoint of the arc $BC$ (not containing $A$) on $(ABC)$. Thus we have $A-I-O$ then note that $AD=AE$ and $\angle DAO=\angle OAE$, thus $\triangle OAD \cong \triangle OAE$ thus we have $\angle ADO=\angle AEO$ and the result follows...thus we are done

C1: I claim that $A$, $I$, and $O$ are collinear. I will prove this by proving that $\angle AIC+\angle OIC=180$. Let $\angle A=2a$, $\angle B=2b$, and $\angle C=2c$. Note that since $O$ is the circumcenter of $\triangle IBC$, we have \[\angle CIO = 90-\frac{1}{2}\angle IOC=90-\angle IBC=90-b.\]Additionally, since $I$ is the incenter of $\triangle ABC$, we have that \[\angle AIC=180-\frac{1}{2}\angle BAC+\frac{1}{2}\angle BCA=180-(a+c)=90+b,\]which is indeed $180-\angle IBC$, meaning that $A$, $I$, and $O$ are collinear, as desired. C2: I claim that $AI$ is the perpendicular bisector of $DE$. This is clearly true, since by tangent properties, we have that $\triangle ADE$ is isosceles with $AD=AE$. C3: Now, note that by (C2), we have that $AI$ is the perpendicular bisector of $DE$. Since by (C1), $A$, $I$ and $O$ are collinear, we also have that $O$ is on the perpendicular bisector of $DE$, meaning that $OD=DE$, implying that $\angle ODE=\angle OED$. This gives that \[\angle ODB = 180-(\angle ADE+\angle ODE)=180-(\angle AED+\angle OED)=\angle OEC,\]which is what we wished to prove, finishing the problem.

Lemma (Incenter-Excenter): $A, I, O$ are collinear. Proof: We prove the stronger statement that $O$ is the intersection of $AI$ and $(ABC)$. Let $O' = AI \cap (ABC)$. Since $ABO'C$ is a cyclic quadrilateral, we have $\angle O'AC = \angle O'BC$ and $\angle O'AB = \angle O'CB$. But $\angle O'AC = \angle O'AB$ because $O'$ lies on the angle bisector of $\angle BAC$, so we have $\angle O'BC = \angle O'CB$, which is equivalent to $O'B=O'C$. Further, we have \begin{align*} \angle O'IB &= \angle O'AB + \angle ABI \\ &= \angle O'BC + \angle IBC \\ &= \angle IBO' \end{align*}which implies that $O'BI$ is an isoceles triangle where $O'I = O'B = O'C$, so $O'$ is the circumcenter of $\Delta BCI$, and $O' = O$. $\blacksquare$ Applying this lemma, we get $\angle OAD = \angle OAE$. Combined with $AD = AE$, we obtain $\Delta OAD = \Delta OAE$ by SAS congruence, implying that $\angle ODA = \angle OEA$. This is equivalent to $\angle ODB = \angle OEC$, as desired. $\blacksquare$

By the Incenter-Excenter Lemma, $A, B, O,$ and $C$ are concyclic, and $A, I,$ and $O$ are collinear. Futhermore, $\triangle DAO$ and $\triangle EAO$ are congruent by $SAS$, since $AD=AE.$ This means that $\angle ADO=\angle AEO,$ so $\angle ODB= \angle OEC.$

We know that $A, I$ and $O$ are collinear (By Incenter-Excenter Lemma). In $\Delta ADO$ and $\Delta AEO$, Since, $\overline{AD} = \overline{AE}$ (Tangent to the incircle) and $\angle DAO = \angle DAI = \angle EAI = \angle EAO$. So, $\Delta ADO \cong \Delta AEO$ (By SAS Congruence Criterion) Hence, $\angle BDO = 180^{\circ} - \angle ODA = 180^{\circ} - \angle OEA = \angle CEO$.

$A$, $I$, and $O$ are collinear and $\angle DAO = \angle EAO$ , $AD = AE$ which means $\Delta ADO$ and $\Delta AEO$ are equal implying the result.

By \textbf{Incenter-Excenter Lemma} we have that $O$ lies on $AI$. We also have $AD=AE$ and $\angle DAO=\angle EAO$. So by SAS congruence criteria $\triangle ADO \cong \triangle AEO\Rightarrow \angle ADO=\angle AEO \Rightarrow \boxed{\angle BDO =\angle CEO} $.

Many people posted solutions to this problem, but I will post mine anyway. I wrote this while working on EGMO.

By Incenter-Excenter Lemma, $O$ is the midpoint of arc $BC$ in $(ABC)$. Since $AD = AE$ and $AI$ is the angle bisector of $\angle A$, then $\triangle ADO \cong \triangle AEO$. Therefore, $\angle ADO \cong \angle AEO$, so $\angle ODB \cong \angle OEC$, as desired.

Rating (MOHs): 0

By Incenter Lemma, $A$, $I$ and $O$ are collinear and $O\in (ABC)$. The result follows by $ADOE$ kite.

By Fact 5 $A,I,O$ collinear. Observe by SAS $\triangle ADO \cong \triangle AEO.$ Then $\angle BDO=\angle DAO+\angle DOI=\angle EAO+\angle EOI=\angle EOC.$

By Fact $5$, $AIO$ is collinear. Also $AD=AE$ by tangents, so $\triangle DAO \cong EAO$ so the exterior $D$ and $E$ angles are equal which finishes.

O is the center of the circle that goes through B, C, I, and by the incenter excenter lemma, it also goes through $I_A$. From there, A, O, I are collinear and the result follows because triangle ADO is congruent to triangle AEO.

Because $I$ is the incenter of $\triangle{ABC}$, we have that $ID = IE$; we also have that $\angle{BAI} = \angle{CAI}$ by properties of the incenter. From the incenter-excenter lemma, we have that $O$ lies on line $AI$. Since $\angle{ADI} = \angle{AEI} = 90^{\circ}$, we have that $\angle{DIO} = \angle{EIO}$. Thus, by Side-Angle-Side congruence (because obviously $IO = IO$), we have that $\triangle{DIO} \cong \triangle{EIO}$. This means that $\angle{IDO} = \angle{IEO} \implies 90^{\circ} - \angle{IDO} = 90^{\circ} - \angle{IEO} \implies \angle{BDO} = \angle{OEC}$, as desired.