Verified answer

Let P, Q, R be the points of contact of AB, BC, CA respectively with the circle, and O be the centre of the circle.

As OR and OQ are radii, and therefore perpendicular to the tangents AC and BC, they are each of length r, making ORCQ a square.

AR = b - r

AP = AR (tangents)

AP = b - r ...(1)

BQ = a - r

BP = BQ (tangents)

BP = a - r ...(2)

Adding (1) and (2):

AP + BP = a + b - 2r

c = a + b - 2r

a + b = c + 2r.