Wernick's list problem number 27: $A,M_a,I$ find vertexes $B$ and $C$

Question

we want to find triangle $\triangle ABC$ with a straightedge and compass and we're given only:

vertex $A$

incenter $I$ and

midpoint of side $a$, $M_a$.

With line segment $AM_a$ I found the centroid $G$ as it divides $AM_a$ in ratio $2:1$.

With the centroid and the incenter $I$ I found the Nagel point but now I'm stuck.

Answer 1

Here a solution: let $P_a$ be the projection of $I$ on the $BC$-side.
$IM_a$ is the hypotenuse of the right triangle $IP_a M_a$ and it is parallel to the line joining $A$ with the Nagel point of $ABC$. This line is also parallel to the line joining $P_a$ with the vertex $A'$ of the anticomplementary triangle, due to the fact that the Nagel and Gergonne points are isotomic conjugates. This observation leads to a very efficient construction:

1. Construct $G$ through $A$ and $M_a$, $A'$ through $A$ and $M_a$
2. Construct the circle with diameter $IM_a$
3. Construct $P_a$ as the intersection between the previous circle and the parallel to $IM_a$ through $A'$
4. The circle centered at $I$ through $P_a$ is the incircle, hence the sides $AB,AC$ can be drawn as tangents from $A$.

Given the freedom you have in point $3.$, in general there are two solutions.