AI.AIa=bc=AIb.AIc
BICIa is cyclic
Suppose extension of AD intersects circumcircle of DIbIc at X
If you draw perpendicular from X to AIb and extend it then it intersects BC at E’
AX=AE’
Suppose extension of AD intersects circumcircle of DIIa at Y
If you draw perpendicular from Y to AIa and extend it then it intersects BC at E’
YN=NE’