Look at picture. I show how to construct midpoints both of arcs $AC$ in $(ABC)$.

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Solution 1(Config ): Extend $AI,BI.CI$ to $M_A,M_B,M_C$, so the same problem with orthocenter Rephrased problem wrote: Given a triangle ABC, its circumcircle with orthocenter $H$, construct its circumcenter Let the foot of altitudes be $D,E,F$ then we can draw the antipode of $A$ by first constructing $EF \cap BC = X_A$ then $AX_A \cap (ABC) = Y_A$ and $Y_AH \cap (ABC)$, we can draw midpoint of $BC$ by $HY_A \cap BC$ , we can also construct the intersection of symmedian with circumcircle by $X_AH' \cap (ABC)=K$ then draw $L$ s.t. $AL \parallel BC$ by $KM \cap (ABC)$. $O = H'L \cap AA'$

math_comb01 wrote: Solution 1: Trivial by Poncelet Steiner Solution 2(Config ): Extend $AI,BI.CI$ to $M_A,M_B,M_C$, so the same problem with orthocenter Rephrased problem wrote: Given a triangle ABC, its circumcircle with orthocenter $H$, construct its circumcenter Let the foot of altitudes be $D,E,F$ then we can draw the antipode of $A$ by first constructing $EF \cap BC = X_A$ then $AX_A \cap (ABC) = Y_A$ and $Y_AH \cap (ABC)$, we can draw midpoint of $BC$ by $HY_A \cap BC$ , we can also construct the intersection of symmedian with circumcircle by $X_AH' \cap (ABC)=K$ then draw $L$ s.t. $AL \parallel BC$ by $KM \cap (ABC)$. $O = H'L \cap AA'$ Hi, how does Solution 1 work? Don't you need the circumcenter already for Poncelet Steiner??

U need to do incenter excenter lemma+relfecction of orthocentre+produce AI ,BI,CI to meet circumcircle again u find I as orthocenter of new triangle

Construct $D=AI \cap BC, E=BI \cap AC, F=CI \cap AB$. Firstly construct $M_A$ as the second intersection between $AI$ and $(ABC)$. Construct $A_1=EF \cap BC$, then $(B,C;D,A_1)=-1$. Since $\angle BAD = \angle CAD$, $AD \perp AA_1$. Construct $M_A'$ as the second intersection between $AA_1$ and $(ABC)$. Since $\angle M_A A M_A' = 90^{\circ}$, $M_AM_A'$ is a diameter of $(ABC)$ so it passes through the circumcenter $O$. Similarly, draw the diameter $M_BM_B'$, and $O=M_AM_A' \cap M_BM_B'$ so we are done.

It's easy to construct tangent on a given point on circle by using the poles(Click to reveal hidden text

). Incenter has already given, hence we can construct a middle of arcs. Line passing throw two these corresponding points also passes throw the circumcenter. Now it's enough to intersect two these lines to get the circumcenter which is exactly what was required.

Let $AI$ meet $(ABC)$ again at the midpoint $D$ of arc $BC$, and construct $E$, $F$ similarly, then $X = EF \cap AI$ is the midpoint of $AI$ (as $EF$ is the perpendicular bisector of $AI$). Construct $Y$, $Z$ similarly. Then $G = BZ \cap CY$ is the centroid of $\Delta IBC$, so $M = IG \cap BC$ is the midpoint of $BC$ and allows us to construct the perpendicular bisector $MD$ of $BC$. Construct the other perpendicular bisectors similarly, then they meet at the circumcenter.