I - the intersection point of the internal bisectors of triangle ABC;
I_b - the center of the circumcircle tangent to side AC and the extensions of sides AB and BC of triangle ABC;
I_c - the center of the circumcircle tangent to side AB and the extensions of sides AC and BC of triangle ABC.
Q - the center of the circumcircle around triangle II_b I_c.
Find the angle between QI and BC.
Angle I_b A I_c has 180, then A is on I_b I_c.
Let M on AB and N on AC such that MN // BC and I is on MN.
angle NII_b=CBI_b=B/2
angle II_cI_b = angle CI_cA = 180-ACI_c-CAI_c = 180-C/2-(A+(B+C)/2)=
=180-C/2-A-B/2-C/2=180-A-(B+2C)/2=(2B+2C)/2-(B+2C)/2=B/2
Therefore angle NII_b = angle II_cI_b , and it results that MN is the tangent in I to the circle C(II_bI_c), so QI is orthogonal to MN, then the angle between QI and BC is 90.
Dinu Teodorescu
Your solution is correct and original! Thank you
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