Notations: $ ABC$ is the triangle, $ G,P,Q$ the given points respectively. $ P,Q$ are opposite $ A$ $ X.Y.Z\implies$ the points $ X,Y,Z$ lies on a line in that order where the order is required. Construction: The following points are easily constructable with the properties with them. $ D: P.D.G\ \&\ PG = 4GD$ $ E: P.Q.E\ \&\ DP = DE$ $ A: AP\perp PQ\ \&\ A.G.E$ $ F: A.Q.F\ \&\ FE\perp PQ$ $ S: SE\perp PQ\ \&\ SA = SF$ $ B,C: SA = SB = SC\ \&\ P.Q.B.C$

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i have another construction i'm not good at constructions, but i think this is easy... i use the same notation from the previous solution... let $ G'$ be the foot of the perpendicular from $ G$ to line $ BC$ (or line $ PQ$, it's the same)... we know that $ GG' = \frac {1}{3}AP$, so we construct $ A$ in the same semiplane of $ G$ wrt line $ BC$, so that $ AP = 3GG'$. then we can construct the midpoint $ M$ of side $ BC$ and let $ l$ be the perpendicular line to $ BC$ passing through $ M$ (i.e, the perpendicular bissector). also, we have that the reflection of $ AP$ across $ AQ$ passes through the circumcenter, so, its intersection with $ l$ gives the circumcenter of triangle $ ABC$. from this, $ B$ and $ C$ are obtained by intersecting this circle and line $ BC$...