at the first sight this problem can be killed by using baricentric coordinates, denote $\overrightarrow{r_{A_{1}}}= x \cdot \overrightarrow{r_{B}}+(1-x) \cdot \overrightarrow{r_{C}}$ etc, and now we have $\overrightarrow{r_{H_{1}}}= \cot{B}\cot{C}r_{A_{1}}...$ etc, because the triangles are similar, and expand $\overrightarrow{r_{H_{1}}}$ and put the condition to be equal with $\overrightarrow{r_{O}}= a \cos{A}\overrightarrow{r_{A}}$ etc, the conclusion follows.

I don't understand your solution. You say that if $\overrightarrow{r_{A_{1}}}= x \cdot \overrightarrow{r_{B}}+(1-x) \cdot \overrightarrow{r_{C}}$ , $\overrightarrow{r_{B_{1}}}= x \cdot \overrightarrow{r_{C}}+(1-x) \cdot \overrightarrow{r_{A}}$, $\overrightarrow{r_{C_{1}}}= x \cdot \overrightarrow{r_{A}}+(1-x) \cdot \overrightarrow{r_{B}}$ then $A_{1}B_{1}C_{1}$ is similar to $ABC$ ? This is not true.

no, i refered just to write $ r_{A_{1}}$ and the analogues as a sum of $ r_{b}, r_{c}$ with constanst $ x,y,z$ not $ x$ in the same place, and put the condition of similarity just when you write $ r_{H_{1}}$.

Invalid URL Here is my proof: Let H be the hortocentre of $A_{1}B_{1}C_{1}$. Since $\angle B_{1}AC_{1}=-\angle B_{1}A_{1}C_{1}= B_{1}HC_{1}$ (using directed angles), it's easy to see that A,B,C are on the circles obtained from the circumcircle of $A_{1}B_{1}C_{1}$ with symmetries on the sides of $A_{1}B_{1}C_{1}$. We need to prove that $HA = HB = HC$. But HB, HC are chords of circles with the same radius. We need only to check that their respective angles on the circumference are equal. But this is indeed very easy! $\angle HA_{1}C = \angle HA_{1}B$, $\angle HC_{1}B = \angle HC_{1}A$, $HB_{1}A = HB_{1}C$ using directed angles! It's also true a nice statament: in the picture, choose arbitrary C on the first circumference, find B as intersection of $CA_{1}$ and the second circumference, find A as intersection of $BC_{1}$ and the third circumference. Then ABC is a triangle, and is a triangle similar to $A_{1}B_{1}C_{1}$.

April wrote: Let $ABC$ is an acute angled triangle and let $A_{1},\, B_{1},\, C_{1}$ are points respectively on $BC,\,CA,\,AB$ such that $\triangle ABC$ is similar to $\triangle A_{1}B_{1}C_{1}.$ Prove that orthocenter of $A_{1}B_{1}C_{1}$ coincides with circumcenter of $ABC$. Proof. Let $O$ be the orthocentre of triangle $A_{1}B_{1}C_{1}$ and let $H=B_{1}O\cap C_{1}A_{1},\,K=C_{1}O\cap A_{1}B_{1}$ We have: $A_{1},\,H,\,O,\,K$ are cyclic. $\Longrightarrow \angle B_{1}OC_{1}=\angle HOK=180^\circ-\angle B_{1}A_{1}C_{1}=180^\circ-\angle BAC$ So the quadrilateral $B_{1}OC_{1}A$ is cyclic. $\Longrightarrow \angle OAC=\angle OC_{1}B_{1}=90^\circ-\angle A_{1}B_{1}C_{1}=90^\circ-\angle ABC$ Similar: $\angle OCA=90^\circ-\angle ABC$ So triangle $OAC$ is isosceles or $OA=OC$ Similar: $OC=OB$ Hence: $O$ is the circumcenter of triangle $ABC$