How to construct the center of an inellipse, if $ABC$ and the major axis are given?

Question

$ABC$ is given, and the major axis. How to construct the center of this inellipse of $ABC$ ?

Answer 1

Suppose $F$ and $G$ are the foci of the required ellipse, inscribed in triangle $ABC$. The following property holds for any conic: equal angles are formed by the tangent lines through an external point $P$ with the lines through $P$ and the foci (see here, for instance). That is: $$ \angle CAF=\angle BAG,\quad \angle ACF=\angle BCG,\quad \angle CBF=\angle ABG. $$ By definition, then, $F$ is the isogonal conjugate of $G$ (and vice-versa) with respect to triangle $ABC$. It follows that the isogonal conjugate of the major axis intersects the major axis itself at $F$ and $G$.

If $DE$ is the given line of the major axis, intersecting sides $AB$, $AC$ of the triangle, then its isogonal conjugate is a hyperbola, passing through vertices $ABC$ and through $D'$, $E'$, isogonal conjugates of $D$, $E$. The hyperbola intersects line $DE$ only if $DE$ intersects segments $IB$ and $IC$, where $I$ is the incenter of $ABC$, which is the isogonal conjugate of itself.

To find the center of the inscribed ellipse we can then find the equation of the hyperbola through $ABCD'E'$ and its interections $F$, $G$ with line $DE$ (if any). Center $O$ is the midpoint of $FG$.