Question 1120565
.
Let me offer different solution.


<pre>
{{{drawing(400,200,-2,22,-2,10,
line(0,0,20,0),line(20,0,20,8),line(20,8,0,8),line(0,8,0,0),line(0,8,4,0),line(4,0,20,8),locate(-1,-.5,A),locate(-1,8.5,B),locate(20.5,8.5,C),locate(20.5,-.5,D),locate(4,-.5,P),
locate(-1,4,"8"),locate(10,9,"20")
)}}}


For right triangles, there is the simple and straightforward formula for the radius of the inscribed circle

    r = {{{(a + b - c)/2}}},

where "a" and "b" are the legs lengths and "c" is the hypotenuse length.

    For the proof of this formula see the lesson 
        <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-tangent-lines-released-from-a-point-outside-a-circle.lesson
>Solved problems on tangent lines released from a point outside a circle</A>
    in this site.

Using this formula, you get

    r1 = {{{(1/2)}}}.(|AB| + |AP| - |BP|)    (1)  for the triangle ABP;

    r2 = {{{(1/2)}}}.(|BP| + |PC| - |BC|)    (2)  for the triangle PBC;  and  

    r3 = {{{(1/2)}}}.(|PD| + |DC| - |PC|)    (3)  for the triangle PDC.


Now add the formulas (1), (2) and (3).  You will get

r1 + r2 + r3 = {{{(1/2)}}}.(|AB| + |AP| - |BP| + |BP| + |PC| - |BC| + |PD| + |DC| - |PC|) = 

                   the terms -|BP|  and  |BP|,  |PC|  and  -|PC|  cancel each other, and you get

             = {{{(1/2)}}}.(|AB| + |AP|  - |BC| + |PD| + |DC|) = {{{(1/2)}}}.(8 + 20 - 20 + 8) = 8.
</pre>

<U>Answer</U>.  &nbsp;&nbsp;r1 + r2 + r3 = 8.