4 concyclic points $S_1$, $S_2$, $T_1$ and $T_2$ are given. Construct a triangle $ABC$ so that its inscribed circle intersects the angle bisector from vertex $A$ at points $S_1$ and $S_2$, and the median from vertex $A$ at points $T_1$ and $T_2$.
I found point A as the intersection point of $S_1S_2$ and $T_1T_2$. Then I found midpoint $J$ of $S_1S_2$ and constructed a circle of radius $|JS_1|$ and center $J$. Finally, I constructed tangents to the circle at point $A$. How do I construct the missing side of $ABC$? I was thinking it would be great if I could just construct the centroid of $ABC$ using its incenter $J$ but I don't know how.
Hint: Suppose tangent points are E and F. Connect E to F. EF intersect $T_1T_2$ at G. Connect G to J. Draw a perpendicular like $l$ on GJ at G. $l$ intersects extensions of AE and AF at point B' and C'. It can be shown that G is mid point of B'C'. Hence BC which is parallel with B'C' and tangent on the circle can be the side you are trying to find. To find that, extend GO from O to meet th circle at H. Draw a perpendicular on OH at H, it intersect AD and AE at B and C, BC is parallel we B'C' and tangent on circle centered at J.