Denote D the midpoint of AB it's easy to show C',D,I are collinear C'D=DI AC'BI is a parallelogram => C' is the orthocenter of △OcAB => Oc,C',H are collinear Analogously Oa,A',H Ob,B',H are collinear ∠HOcB=∠HOaB=1/2∠ABC => HOa=HOc Analogously HOa=HOc=HOb H is the circumcenter of △OaObOc

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We denote by $S$ the incentre of the medial triangle and by $I,G,H,O$ the incentre,centroid,orthocentre,circumcentre of $\triangle ABC$ respectively.Now we know that $I$ is the orthocentre of the excentric triangle.This can be easily shown by using the fact that the internal bisectors are perpendicular with the external ones and the excentres are collinear with the incentre and the corresponding vertices Hence $\triangle ABC$ is the orthic triangle of its excentric triangle,which implies that $O$ is the nine point centre of the excentric triangle.Hence the circumcentre of the excentric triangle is the reflection of $I$ over $O$,say $P$. Now $\triangle A'B'C'$ is homothetic with $\triangle ABC$ with centre $S$ and ratio 1.Hence the orthocentre of $\triangle A'B'C'$ is the reflection of $H$ over $S$,say $Q$.We have to show that $P$ and $Q$ are the same point. Now the medial triangle is homothetic with $\triangle ABC$ with centre $G$ and ratio $\frac{-1}{2}$.Hence $I,G,S$ are collinear and $\frac{IG}{GS}=2$.Also we know that $\frac{HG}{OG}=2$ and $G$ lies on $OH$.Hence in triangles $OGS$ and $HGI$ we have \[\frac{IG}{GS}=\frac{HG}{GO}=2\].Hence these two triangles are similar,which implies $OS\parallel IH$ and$2OS=IH$.Now let $IO$ and $HS$ intersect at $R$.Now from the previous relation we easily get that $IO=OR$ and $HS=SR$,which implies that $P,Q,R$ all are the same points,which was to be shown.

This is just restatement of the well known result: Quote: In any triangle, the Spieker point is the midpoint of the orthocenter and the Bevan point. See Episodes in Nineteenth and Twentieth Century Euclidean Geometry , Ross Honsberger for a proof.