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Construct $ABC$ with straightedge and compass, given $G,I,Q_a$.

$G - $ the intersection point of medians;

$I -$ the center inscribed circle;

$Q_a -$ point of tangent inscribed circle to the side of $BC$.

I think $IQ_a \perp BC$, but I don't know the properties that combines these three points.

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An interesting approach may be the following one: you may construct the Nagel point $N$ as $3G-2I$. You have the $BC$-line, i.e. the perpendicular to $IQ_a$ through $Q_a$.
You may assume that some $P_a\in BC$ is the contact point of the $A$-excircle, then:

  1. $A$ lies on $NP_a$;
  2. The midpoint of $M_a$ of $P_a Q_a$ is also the midpoint of $BC$;
  3. $A$ lies on $M_a G$ (so $A$ is fixed by $P_a$ through $A=NP_a\cap M_a G$);
  4. Assume that the perpendicular to $BC$ through $P_a$ meets the $AI$-line at $J_a$: if the distance of $J_a$ from $BC$ is the same as the distance of $J_a$ from $AB$ then $J_a$ is the $A$-excenter and the problem is solved.

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Now $J_a$ lies on a hyperbola, while the external angle bisectors of the candidate $B,C$ vertices meet on a line, so the problem boils down to intersecting a line and a hyperbola, finding the actual $A$-excenter, then the contact point of the $A$-excircle, then the midpoint of $BC$, then $A$ as $NP_a\cap M_a G$, then $B$ and $C$ by drawing the tangents from $A$ to the incircle.

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  • $\begingroup$ I do not understand how to find $J_A$. Why $J_{a}$ lies on a hyperbola? $\endgroup$
    – Roman83
    Commented Jun 25, 2016 at 17:16
  • $\begingroup$ @Roman83: because the map sending $J_a$ to $P_a$ is a projectivity. If you choose a coordinate system, that is straightforward to check. So the idea is to pick five differents $P_a$, reconstruct through them the foci of the hyperbola, then intersect the hyperbola with the appropriate line. $\endgroup$ Commented Jun 25, 2016 at 17:29

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