Let ABC is a right angle triangle and r is the radius of the circle inscribed in the triangle.

Now, we have to prove that:

r = Perimeter of the ΔABC/2 - Hypotenuse of the triangle

i.e. r = Perimeter of the ΔABC/2 - AC

Now, from the figure,

OP = OQ = r

Again in quadrilateral OPBQ

∠B = 90, ∠P = 90   {since PB is a tangent and OP is redius}

∠Q = 90,  {since BQ is a tangent and OQ is redius}

Since 3 angles are right angle,

So, ∠O is also right angle

i.e. ∠O = 90

Again, adjacent sides OP and OQ are equal.

Hence, quadrilateral OPBQ is a square.

Hence, PB = BQ = r

Now, perimeter of the ΔABC = AB + BC + AC

=> Perimeter of the ΔABC = (AP + PB) + (BQ + QC) + (AR + RC)

Since, AP = AR, PB = BQ, QC = CR   {tangent drawn from external point to the circle are equal}

=> Perimeter of the ΔABC = AP + r + r + QC + AR + RC

=> Perimeter of the ΔABC = AP + QC + AR + RC + 2r

=> Perimeter of the ΔABC = AR + RC + AR + RC + 2r

=> Perimeter of the ΔABC = 2AR + 2RC + 2r

=> Perimeter of the ΔABC = 2(AR + RC) + 2r

=> Perimeter of the ΔABC = 2AC + 2r

=> Perimeter of the ΔABC = 2(AC + r)

=> Perimeter of the ΔABC/2 = AC + r

=> r = Perimeter of the ΔABC/2 - AC

Hence proved.