Proving circumcenter lies on altitude

Problem: In $\triangle ABC$, let $D$ be the intersection of the tangents to the circumcircle at $B$ and $C$, let $B'$ be the reflection of $B$ across $AC$, let $C'$ be the reflection of $C$ across $AB$. Prove that the circumcenter of $\triangle DB'C'$ lies on the altitude from $A$ in $\triangle ABC$.

If we let $O_2,O_3$ be the circumcircles of $\triangle AB'C,\triangle AC'B$, then notice that $\triangle C'O_3O\cong\triangle B'O_2O$. This is definitely true, because it is necessary for the problem statement to be true (also I tested it on GeoGebra): Clearly $H\in (AB'C),(AB'C)$, where $H$ is the orthocenter of $\triangle ABC$ (since $\angle AHB=\beta+\gamma,\angle AB'C=\angle AC'B=\alpha$, where $\angle BAC=\alpha,\angle ABC=\beta, \angle BCA=\gamma$). Then since circles $(AC'B),(AB'C),(ABC)$ are all congruent, by the congruence of the triangles, we have $OO_2=OO_3\implies \text{pow}_{(AC'B)}(O)=\text{pow}_{(AB'C)}(O)$. But since $AH$ is the radical axis of $(AC'B),(AB'C)$, $O\in AH$, so $O$ lies on the altitude from $A$.

But I couldn't manage to prove that $\triangle C'O_3O\cong\triangle B'O_2O$. Clearly $OC'=OB',O_3C'=O_2B'$, but I couldn't manage to prove that $\angle OC'O_3=\angle OB'O_2$ (though I know it's definitely true).

• My apologies, I was going to add it after I typed everything but forgot. It's up now. – Max Jul 19 '16 at 15:37

You may prove through angle chasing or the cosine theorem that $DC'=DB'$.
We have $\widehat{C'AB'}=3\widehat{A}$ and $\widehat{C'BD}=\widehat{B'CD}=\pi-(\widehat{B}-\widehat{C})$: notice that its supplementary angle is exactly the angle between $AH$ and $AO$, if $H$ and $O$ are the orthocenter and circumcenter of $ABC$ (they are isogonal conjugates). Moreover, $\widehat{B'DC'}=2\widehat{A}$ and $CDB$, $B'DC'$ are similar triangles. By Thales' theorem, the circumcenter of $DCB$ is just the midpoint of $OD$.
We may compute $B'C'^2$ by applying the cosine theorem to $AB'C'$ and get: $$B'C'^2 = b^2+c^2-2bc\cos(3\widehat{A})$$ so the circumradius of $B'C'D$ is straightforward to compute through $B'C'$ and $\widehat{B'DC'}$.
If $U$ is the circumcenter of $B'C'D$, $\widehat{B'UC'}=2\widehat{B'D C'}=4\widehat{A}$.