The distance from the center (the intersection of three bisectors) to three sides is equal (theorem, which can also be proved by bisector theorem)

Let the intersection of bisectors be 0, and bisectors with angles < a < b < c intersect BC, AC and AB at D, E and F..

The area of the triangle S=AC*BC*( 1\2)= the area of the triangle BOC+BOA+ACO (because the heights of the three triangles are equal, the distance is set to h).

That is, AC * BC = (AB+BC+AC) * h.

The solution is h= 1.