Given triangle abc with angle abc congruent to angle acb, which step could be used to prove that side ab is congruent to side ac?

From A draw the altitude AH intersecting BC in H

Let's prove that triangle ABH is congruent to triangle ACH

The above 2 triangles are right triangles due to the altitude AH

Angle B= Angle C (given)
Angle AHB = Angle AHC =90° (since AH is the altitude)
Then angle BAH = CAH (both complementary to B & C respectively
And AH is a common side
Now Tri. ABH = Tri. ACH because ASA, hence AB=AC