Construct a triangle given its circumscribed circle and the three points on it...

Question

Construct a triangle given its circumscribed circle and the three points on it at which the altitude, the angle bisector and the median, drawn from the same vertex, intersect the circle.

The problem is from the book. I have copied the question exactly how the book has stated it including punctuations. Below is something I made as a quick "glance" of sorts. Red figures are the givens. .

Answer 1

Let $H_A,L_A,M_A$ be the points on the circumcircle given by the altitude, the angle bisector and the median through $A$. Let $O$ be the center of the circumcircle (i.e. the circumcenter of $ABC$).

1. The parallel to $O L_A$ through $H_A$ meets the circumcircle at $A$;
2. $AM_A$ and $OL_A$ meet at the midpoint of $BC$;
3. The perpendicular to $OL_A$ through the midpoint of $BC$ is the $BC$ side.

If you do not know which point lies on which cevian you have to consider that on the $BC$ side the foot of the angle bisector always lies between the foot of the altitude and the midpoint of the $BC$ side by the bisector theorem. It follows that there are at most two solutions associated with the labelings $(H_A,L_A,M_A)$ and $(M_A,L_A,H_A)$: