<Tri-2024-2157>, hint

Let N be  a point on line CA such that AI=AN and M be intersection of BN with line CI

then M is the midpoint of BN. Since triangles IBM and INM are congruent

<BIM=<NIM=90-A/2 <IAB=<INB=<A/2  Hence IANB are cyclic

<A=<ANB+<ABN=<ANI+<INB+<ABN=<B/2+<A/2+<B/2

Hence <A=2<B.