Solution:

Claim 1. It is possible to draw internal/external angle bisectors.
Proof. Let $A$, $B$, $C$ be marked. To bisect $\angle A B C$, draw the parallel line to $A B$ unit 1 away from it on the opposite side as $C$, and draw the parallel line to $B C$ unit 1 away from it on the opposite side as $A$. Let these lines intersect at $D$. Then $B D$ is the internal angle bisector of $\angle A B C$. We can construct external angle bisectors similarly by drawing the line on the same side as $A$ for the second line instead.

Corollary 2. It is possible to mark the incenters and excenters of a triangle $A B C$.
Proof. Draw in the internal/external bisectors of all three angles and intersect them.

Claim 3. It is possible to mark the midpoint of any segment $A B$.
Proof. Let $B$ and $C$ be marked. Draw an arbitrary point $A$ not on line $B C$. Draw a line parallel to $B C$ unit 1 away from it on the opposite side as $A$, and let this line intersect $A B$ at $D$ and $A C$ at $E$. Let $B E$ and $C D$ intersect at $F$, and let $A F$ intersect $B C$ at $M$. Then by Ceva's Theorem, $M$ is the midpoint of $B C$.

Corollary 4. It is possible to mark the centroid of $A B C$.
Proof. Draw the midpoint $D$ of $B C$ and the midpoint $E$ of $A C$, and intersect $A D$ with $B E$.

Claim 5. It is possible to draw the perpendicular bisector of any segment $B C$.
Proof. Let $B$ and $C$ be marked. Draw an arbitrary point $A$ not on line $B C$. Construct the incenter $I$ and $A$-excenter $I_A$ of $A B C$. Draw the midpoint $M$ of $B C$ and midpoint $N$ of $I I_A$. By the incenter-excenter lemma, $N$ is the midpoint of the $\text{arc B C}$ not containing $A$, so $M N$ is the perpendicular bisector of $B C$.

Corollary 6. It is possible to mark the circumcenter of $A B C$.
Proof. Draw and intersect the perpendicular bisectors of $B C$ and $A C$.

Claim 7. Given two marked points $A$ and $B$, it is possible to mark the point $C$ such that $\overrightarrow{B C}=\frac{1}{2} \overrightarrow{A B}$.
Proof. Draw an arbitrary point $D$ not on line $A B$. Draw the midpoint $M$ of $A D$. Draw the midpoint $M_1$ of $B D$ and the midpoint $M_2$ of $M D$, and let $M_1 M_2$ intersect $A B$ at $C$. Then $M_1 M_2 \parallel B M$ and $M M_2=\frac{1}{2} M D=\frac{1}{2} A M$, so $B C=\frac{1}{2} A B$.

Claim 8. Given two marked points $A$ and $B$ and any positive real number $k$ such that $2 k \in \mathbb{Z}$, it is possible to mark the point $C$ such that $\overrightarrow{B C}=k \overrightarrow{A B}$.
Proof. Note that by applying Claim 7 and marking the midpoint of $A B$, we can translate both $A$ and $B$ by $\frac{1}{2} \overrightarrow{A B}$. The claim now follows by applying this operation repeatedly.

To finish, take the given triangle $A B C$ and mark its circumcenter $O$ and centroid $G$. Note that its orthocenter $H$ satisfies that $\overrightarrow{G H}=2 \overrightarrow{O G}$, so applying Claim 8 to $k=2$ now finishes the problem.

Solution:

Start with Claims 1-3 of solution 1, allowing us to draw internal/external angle bisectors, in/excentres, and midpoints. We add one more claim.

Claim 9. Given a point $P$ and a line $\ell_1$, it is possible to draw a line through $P$ parallel to $\ell_1$.
Proof. Draw the line $\ell_2$ on the opposite side of $\ell_1$ to $P$, a distance 1 away. Draw arbitrary lines $P A B$ and $P C D$ with $A, C \in \ell_1$, $B, D \in \ell_2$. Let $E$ be the midpoint of $A C$, let $F=P E \cap \ell_2$, and let $Q=B E \cap F C$.

Since $\triangle Q E C \sim \triangle Q B F$ and $\triangle P A E \sim \triangle P B F$, we have
$$\frac{Q E}{Q B}=\frac{E C}{B F}=\frac{A E}{B F}=\frac{P A}{P B}$$
so $\triangle B A E \sim \triangle B P Q$. In particular, $P Q$ is parallel to $A E$, as desired.

In triangle $\triangle A B C$, draw the incentre $I$ and $A$-excentre $I_A$. Draw the midpoints $D$ of $B C$ and $M$ of $I I_A$. By the incentre-excentre lemma, $M$ is on the perpendicular bisector of $B C$, so $M D$ is perpendicular to $B C$. Finally, using Claim 9, we can draw a line through $A$ that is perpendicular to $B C$. Repeat this for $B$ and $A C$, and their intersection is the orthocentre of $\triangle A B C$, as required.