This is more a comment than a solution, using a routine method—the $p,q$-method—to reduce the problem by means of simple calculations to that of determining whether the determinant of a $6\times 6$ matrix vanishes.

We can assume that the triangle vertices are $(0,0),(1,0),(p,q)$—then the intermediary points are $D=(\lambda,0)$, $E=(1-\lambda+\lambda p,\lambda q)$ and $F=(\lambda p,\lambda q)$ with $\lambda=\frac 1 k$.

It is then routine to compute the unit normals to $DE,EF,FD$ and so $|IB|,|CG|,|AH|$ in terms of $\lambda,p,q$. This gives the coordinates of your six points and one can then use the fact that they lie on a conic if and only if the determinant of the $6\times 6$ matrix with rows $$ [x^2\ xy\ y^2\ x\ y\ 1] $$ as $(x,y)$ ranges over the coordinates of the points vanishes.

Up to this point, the calculations can be done by hand but that of the determinant is probably intractable and can be carried out using Mathematica.

An advantage of this approach is that it is easy to add parameters to explore possible generalisations or converses. Suggestions: replace $\lambda$ by three distict parameters to determine under which conditions on them the result is still valid; replace the equality conditions by proportionality ones; replace the perpendicular distance by the lengths of lines which meet the sides of $DEF$ at a fixed angle.

This is more a comment than a solution, using a routine method—the $p,q$-method—to reduce the problem by means of simple calculations to that of determining whether the determinant of a $6\times 6$ matrix vanishes.

We can assume that the triangle vertices are $(0,0),(1,0),(p,q)$—then the intermediary points are $D=(\lambda,0)$, $E=(1-\lambda+\lambda p,\lambda q)$ and $F=(\lambda p,\lambda q)$ with $\lambda=\frac 1 k$.

It is then routine to compute the unit normals to $DE,EF,FD$ and so $|IB|,|CG|,|AH|$ in terms of $\lambda,p,q$. This gives the coordinates of your six points $J,K,M,N,O,P$ and one can then use the fact that they lie on a conic if and only if the determinant of the $6\times 6$ matrix with rows $$ [x^2\ xy\ y^2\ x\ y\ 1] $$ as $(x,y)$ ranges over the coordinates of the points vanishes.

Up to this point, the calculations can be done by hand but that of the determinant is probably intractable and can be carried out using Mathematica.

An advantage of this approach is that it is easy to add parameters to explore possible generalisations or converses. Suggestions: replace $\lambda$ by three distict parameters to determine under which conditions on them the result is still valid; replace the equality conditions $|FJ|=|BI|$, etc., by proportionality ones $|FG|=\rho|BI|$ ; replace the perpendicular distance by the lengths of lines which meet the sides of $DEF$ at a fixed angle.

There is a routine method to solve this and similar problems by calculation—the $p.q$-method. We can assume that the triangle vertices are $(0,0),(1,0),(p,q)$—then the intermediary points are $(\lambda,0),(1-\lambda+\lambda p,\lambda q),(\lambda q)$ with $\lambda=\frac 1 k$.

It is then routine to compute the unit normals to $DE,EF,FD$ and so $|IB|,|CG|,|AH|$ in terms of $\lambda,p,q$. This gives the coordinates of your six points and one can then use the $6\times 6$ determinental condition to verify that they lie on a conic. This can be done in principle by hand but this is probably a case where Mathematica is your friend.

An advantage of this approach is that it is easy to add parameters to explore possible generalisations or converses.

This is more a comment than a solution, using a routine method—the $p,q$-method—to reduce the problem by means of simple calculations to that of determining whether the determinant of a $6\times 6$ matrix vanishes.

We can assume that the triangle vertices are $(0,0),(1,0),(p,q)$—then the intermediary points are $D=(\lambda,0)$, $E=(1-\lambda+\lambda p,\lambda q)$ and $F=(\lambda p,\lambda q)$ with $\lambda=\frac 1 k$.

It is then routine to compute the unit normals to $DE,EF,FD$ and so $|IB|,|CG|,|AH|$ in terms of $\lambda,p,q$. This gives the coordinates of your six points and one can then use the fact that they lie on a conic if and only if the determinant of the $6\times 6$ matrix with rows $$ [x^2\ xy\ y^2\ x\ y\ 1] $$ as $(x,y)$ ranges over the coordinates of the points vanishes.

Up to this point, the calculations can be done by hand but that of the determinant is probably intractable and can be carried out using Mathematica.

An advantage of this approach is that it is easy to add parameters to explore possible generalisations or converses. Suggestions: replace $\lambda$ by three distict parameters to determine under which conditions on them the result is still valid; replace the equality conditions by proportionality ones; replace the perpendicular distance by the lengths of lines which meet the sides of $DEF$ at a fixed angle.