By radical axis $A_1C_1,AC,B_0Q$ are concurrent and hence by projecting $B_0(A,C;BB_0\cap AC,B_0Q\cap AC)=-1$ we obtain the desired.

This was also Sharygin 2016. Why did they reuse a problem?

Dear Mathlinkers, can some one explain the proof above??? Is it correct? Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, can some one explain the proof above??? Is it correct? Sincerely Jean-Louis $ACA_1C_1$ is cyclic as is $A_1C_1B_0Q$,so by using radical axis on the last $2$ cyclic quads and $\odot ABC$ we have that $B_0Q,AC,A_1C_1$.Let $BB_0\cap AC=\{B_1\}$ $\implies$ $(A,C;B_1,AC\cap A_1C_1)=-1$ so from projecting we're done.

Dear, thank for your answer... What I have not understand is that when you say "so from projecting we're done." Can you explain more because I have a doubt... Sincerely Jean-Louis

Dear, if we consider the circle (B°B1Q), it goes through the midpoint of BC... The rest... Sincerely Jean-Louis

jayme wrote: Dear, thank for your answer... What I have not understand is that when you say "so from projecting we're done." Can you explain more because I have a doubt... Sincerely Jean-Louis Well okay.We have $(A,C;B_1,AC\cap A_1C_1)=-1$.Project this from $B_0$ to $\odot ABC$ $A,C$ go to $A,C$ and $B_0B_1\cap \odot ABC=\{B\}$ and hence $B_1$ goes to $B$.As showed previously,from radical axis argument $AC\cap A_1C_1$ goes to $Q$ so $(A,C;B,Q)=-1$ $\implies$ $BQ$ is the symmedian.

Dear, sorry to disturb you again...A agree with your explication until (A, C, B, Q) = -1 but why BQ is then the symmedian? Perhaps I am a little blind to day... tahk for your patience... Sincerely Jean-Louis

No problem.Well $AA,CC,BQ$ are concurrent by definition of symmedian $\implies$ $BACQ$ is harmonic.

Dear, Thank you very much... I have totaly forgotten that BACQ was harmonic...I was very blind to day... Very sincerely Jean-Louis

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%206.pdf p. 27... Sincerely Jean-Louis

Solving this because point names are a bit weird. \section{Sharygin Geometry Olympiad First Round 2017/12} Quote: Let $\triangle ABC$ be a triangle with $\triangle DEF$ as its orthic triangle and $H$ as its orthocenter. If $H'$ is reflection of $H$ over $D$, then prove that $(H'EF)$ cuts $(ABC)$ at point $K$ such that $\overline{AK}$ is the $A$-symmedian. Super well known, but just for fun we are going to do this by Ratio Lemma. Define \[f(\bullet)=\pm \frac{B \bullet}{C \bullet}\]where the ratio is negative iff $\bullet$ lies on segment $\overline{BC}$ or under that line. And now $(BEFC)$ is cyclic so by well known lemma which is consequence of ratio lemma we just need to show that \[f(E)f(F)=f(H')f(K)=-f(H')f(A)=-f(D)\]But now see that by Law of sines we have \[f(E)f(F)=\frac{\tan C}{\tan B}=\frac{BA}{AC} \cdot \frac{\cos B}{\cos C}=\frac{BD}{CD}=-f(D)\]And done.