Claim: Given $BC$ we can construct its midpoint.
Proof: Let $\ell$ be a line parallel and $1$ away from $BC$ . After choosing arbitary points $D,E$ on $\ell$ we consider line passin through $BD\cap CE$ and $BE\cap CD$ to bisect $BC$ . $\square$

Using this we can construct the centroid of $ABC$ .

Claim: We can construct the angle bisector of $ABC$ .
Proof: Construct the lines parallel and $1$ away to $AB$ and $BC$ , the line connecting intersection of these two parallel lines and $B$ marks the angle bisector. $\square$

Using this we can construct the incenter and excenters of $ABC$ . And due to INMO 2025 P3 we can construct the perpendicular bisector of $BC$ and get the circumcenter of $ABC$ as well.

Claim: Given $BC$ , we can construct the reflection of $B$ over $C$ .
Proof: Let $\ell$ be a line parallel to $BC$ , then choose $2$ points $D,E$ on $\ell$ and consider their midpoint $N$ . Then consider $BD\cap CN$ to be $R$ . And finally $RE\cap BC$ gives us the relfection of $B$ over $C$ . $\square$
(this claim is basically just working backwards on BPT such that we start with only $2$ points.)

Since we can already construct $O$ , $G$ we can also construct $H$ by noting that $HG=2GO$ and using the above Claim two times.