Given conditions are that, in a triangle ABC, ∠C=90°, and the length of each side is AB=c, BC=a, AC=b, and the radius of inscribed circle is r. Let the center of the circle be O, and the foot of perpendicular from O to each side AB, BC, AC be L, M, N respectively.
The definition of a inscribed circle in a triangle says that the circle comes into contact with each side at a point, but doesn't intersect them. Therefore, the three sides are tangents of the circle, and those tangents are perpendicular to the radii at the points of contact.
That is: L,M,N are the very points of contact, so OL=OM=ON=r, and OL⊥AB, OM⊥BC, ON⊥AC.
From those facts mentioned above, the quadrilateral OMCN is a quare (∠OMC=∠ONC=∠C=90°) and OM=ON=CM=CN=r, and ΔAOL≡ΔOAN (RHS), ΔOBL≡ΔOBM(RHS).
Thus we have: AL=AN=AC-CN=b-r, and BL=BM=BC-CM=a-r.
While, AB=AL+BL, so that c=(b-r)+(a-r)=a+b-2r. Therefore, r=(a+b-c)/2 QED.