Given a right triangle with legs a and b and hypotenuse c, we are asked to prove that a+b=c+d, where d is the diameter of the incircle (inscribed circle.)

Let r be the radius of the incircle. By definition, the center of the incircle lies on the intersection of the angle bisectors of the triangle. A point on an angle bisector is equidistant from the sides of the angle. Thus the radius of the incircle is the perpendicular distance from the center to the sides of the triangle.

Consider the area of the triangle. If the center of the incircle is denoted O, then we can divide `Delta ABC` into three triangles `Delta AOB, Delta BOC, Delta AOC.` The area of each of these triangles is one half the product of the radius and the corresponding side length.

So the area of `Delta ABC` can be written as follows:

`1/2ra+1/2rb+1/2rc=1/2r(a+b+c)` so `r=(2A)/(a+b+c)` where A is the area of `Delta ABC`. Since d=2r, we have `d=(4A)/(a+b+c)`

Now we can find A, the area of `Delta ABC`, using the legs, since it is a right triangle. So `A=1/2ab` . Now `d=(4A)/(a+b+c)=(4(1/2ab))/(a+b+c)=(2ab)/(a+b+c)`

Now consider c+d. `c+(2ab)/(a+b+c)=(ac+bc+c^2+2ab)/(a+b+c)`

We have a right triangle `c^2=a^2+b^2` so

`c+d=(ac+bc+a^2+b^2+2ab)/(a+b+c)`

`=(ac+bc+(a+b)^2)/(a+b+c)`

`=((a+b)^2+(a+b)c)/(a+b+c)`

`=((a+b)(a+b+c))/(a+b+c)`

=a+b QED.

See the image attached.

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