Construct triangle given the circumcenter, incenter and the intersection point of the extension of one of the bisectors with the circumscribed circle.

Question

Kiselev Planimetry ex.596. The question is the one on the title. Here is my progress so far:

Let $C$ be the circumcenter, $I$ the incenter and $D$ be the intersection point of the extension of one of the bisectors with the circumscribed circle.

1. Form the circumscribed circle (since we have $C, D$).
2. Connect $D, I$ with a line, find the intersection point with the circumscribed circle. This is a vertex of the triangle, call it $A$.
3. $OD$ is perpendicular to $BC$, since the arcs $BD$, $DC$ have the same length.

Given the above, we want to find a point (which will be $B$) that a) will lie on the arc $AD$, and b) $BI$ bisects the angle formed by BA, and the perpendicular to $OD$, dropped from $B$.

Can one provide a hint to continue from here, or any further idea? Essentially this question has been asked before here 1, but there was no (correct) answer.

Answer 1

Hints:

Apply Euker's theorem:

$d^2=R(R-2r)$

where d is distance between I (incenter) and O(circumcenter), R=OD is the radius of circle and r is the radius of inscribed circle.You foud vertex A Now draw inscribed circle. Draw two tangents from A on it. The intersection of these tangents with circle give B and C.