CLAIM 1: We can construct foot of perpendiculars (and orthocentre) PROOF: Reflect $A$ in $BC$ to get $A’$. Foot of $A$-perpendicular will be $AA’ \cap BC=D$ Similarly we can construct $E,F$ as feet of $B$ and $C$-perpendiculars. Orthocentre $H$ is just the intersection of the altitudes Now reflect $A$ in $EF$ to get $AA_{1}$. Since $ \triangle AEF \sim \triangle ABC$, we get $\angle EAA_{1}=\angle BAD \implies AA_{1}$ passes through circumcentre $O$. Similarly construct $BB_{1}$ and $CC_{1}$. $O$ is just their intersection, and we are done. $\blacksquare$