In a triangle with sides a, b, c, define s = (a+b+c)/2. If the incircle touches the three sides at points dividing them into segments x, y, z (with x from the vertex opposite side a, etc.), which expression gives x?

1. The tangent lengths from each vertex to the incircle are equal: x, y, z. 2. Side a opposite vertex A has length (y + z), side b = (z + x), side c = (x + y). 3. Adding: a + b + c = 2(x + y + z), so x + y + z = s. 4. Subtracting a = y + z from s gives x = s - a. 5. So x = s - a, by similar argument y = s - b, z = s - c. _Source: SCERT Kerala Std X Mathematics Part-2, Chapter 7-8 (pp 180-199, 2019 ed.)._