am I missing something or it should be like this: $B'$ is the midpoint of $AOB$... :

Nice problem. I think $B'$ is midpoint of $AOC$ and it is ok. Let $X$ and $Y$ be the points on lines $OB$ and $OC$ respectively s.t. $OX=OY=OA$. Let $P$ an $Q$ be points s.t. $AXP \sim AOC$ and $AYQ \sim AOB$. Let $D$ be a symetric point to $A$ w.r.t. $XY$. We have $AC'B \sim AOX$ so $ABX \sim AC'O$ and $AXP \sim AOC$. Thus $AXPB \sim AOCC'$ and $OCC' \sim XPB$. $O$ is circumcenter of $AXY$ so $\angle AXP=\angle AOC=2\angle AXY=\angle AXD$ so $P$ lies on $XD$. We also have $\frac{XP}{XD}=\frac{XP}{XA}=\frac{OC}{OA}=\frac{XB}{XO}$ so $OCC' \sim XPB \sim XDO$ and $OBB' \sim YDO$. Thus $\angle BB'O+\angle CC'O=\angle DOY+\angle DOX=\angle BOC$ so cicumcircles $COC'$ and $BOB'$ are tangent.

Omid Hatami wrote: $O$ is a point inside triangle $ABC$ such that $OA=OB+OC$. Suppose $B',C'$ be midpoints of arcs $\overarc{AOC}$ and $AOB$. Prove that circumcircles $COC'$ and $BOB'$ are tangent to each other. use inversion around $O$ the problem will be: Problem.in triangle $ABC$ let $O$ be a point that $\frac{1}{OA}=\frac{1}{OB}+\frac 1{OC}$ let ex-anglebisector of $\angle AOB$ intersect $AB$ at $C'$ and $B'$ be intersection point of ex-anglebisector of $\angle AOC$ with $AC$.prove that $CC'\parallel BB'$ proof.as problem said we have $OB.OC=OA.OC+OA.OB\Longrightarrow \frac{OA}{OB}=\frac{OC-OA}{OC}$ also we have $\frac{OA}{OB}=\frac{AC'}{BC'}\ ,\ \ \frac{OA}{OC}=\frac{AB'}{CB'}\Longrightarrow \frac{OC-OA}{OC}=\frac{AC}{CB'}$ hence $\frac{AC'}{BC'}=\frac{AC}{CB'}$ hence $CC'\parallel BB'$.

Amir.S wrote: use inversion around $O$ the problem will be: Problem.in triangle $ABC$ let $O$ be a point that $\frac{1}{OA}=\frac{1}{OB}+\frac 1{OC}$ let ex-anglebisector of $\angle AOB$ intersect $AB$ at $C'$ and $B'$ be intersection point of ex-anglebisector of $\angle AOC$ with $AC$.prove that $CC'\parallel BB'$ I can't understand how this happens(it doesn't seem obvious(at least for me))?

Uh sorry i disunderstood problem from @Babak Chalebi's post, thinking that B' is the midpoint of AC.Sorry

Lemma. Given triangle $ABC$, bisectors of angle $B$ and $C$ meet $AC$ and $AB$ at $E,F$, respectively. $P$ is an arbitrary point on $EF$. Let $XYZ$ be pedal triangle of $P$ wrt $\triangle ABC$. Then $PX=PY+PZ.$ This lemma is well-known so I leave to the readers. Back to our problem. Let $O_a, O_b, O_c$ be the circumcenters of triangles $BOC, COA, AOB$, $I_b, I_c$ be the circumcenters of triangles $BOY, COZ.$, $M,N,P$ be the midpoints of $OA, OB, OC,$ respectively. Since $I_bO_b\perp OY, O_bO_c\perp OA$ then $\angle I_bO_bO_c=\angle YOM.$ Similarly, $\angle I_bO_bO_a=180^\circ-\angle YOC.$ $Y$ is the midpoint of arc $AOC$ then $OY$ is the external bisector of $\angle AOC.$ We get $\angle I_bO_bO_c=\angle I_bO_bO_a$ or $I_b$ is the intersection of the bisector of angle $O_b$ and $O_aO_c.$ Similarly, $I_c$ is the intersection of the bisector of angle $O_c$ and $O_aO_b.$ But $OA=OB+OC$ hence $OM=ON+OP$, this means the distance from $O$ to $O_bO_c$ is equal to the sum of distances from $O$ to $O_aO_b$ and $O_aO_c$. According to lemma above, $O$ lies on $I_bI_c$. In other words, $(I_b)$ is tangent to $(I_c)$ at $O$.

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Sorry to revive, but I'm not familiar with the above lemma about pedal triangles. Can someone share a proof or some references concerning the subject?

$O$ is a point inside triangle $ABC$ such that $OA=OB+OC$. Suppose $B',C'$ be midpoints of arcs $AOC$ and $AOB$. Prove that circumcircles $COC'$ and $BOB'$ are tangent to each other.

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