If a, b, c are the sides of a right triangle where c is the hypotenuse,prove that the radius r of the circle which touches the sides of the triangle is given by2a b cr + −

The question seems to be asking for a proof that the radius (r) of the incircle (circle inscribed within the triangle) of a right triangle with sides a, b, and hypotenuse c is given by the formula r = (a+b-c)/2. Here's a step-by-step proof:

Step 1: The area of a triangle can be given by the formula A = 1/2 * base * height. For a right triangle, this becomes A = 1/2 * a * b.

Step 2: The area of a triangle can also be given by the formula A = r * s, where r is the radius of the incircle and s is the semi-perimeter of the triangle.

Step 3: The semi-perimeter of a triangle with sides a, b, and c is given by s = (a+b+c)/2.

Step 4: Substituting the semi-perimeter into the formula from Step 2, we get A = r * (a+b+c)/2.

Step 5: Setting the two formulas for the area of the triangle equal to each other, we get 1/2 * a * b = r * (a+b+c)/2.

Step 6: Solving for r, we get r = (a+b-c)/2.

So, the radius of the incircle of a right triangle with sides a, b, and hypotenuse c is given by r = (a+b-c)/2.