CPhill

Here's 4

1.   angle 1 + angle 2  = 90                                                     given,  defintion of complementary

2.   angle 3 + angle 2  = 90                                                    given,  definition of  complementary

3.   angle 1 + angle 2  = angle 3 + angle 2                             equating 1, 2 with substitution

4.   angle 1 = angle 3                                                              subtract angle 2 from both sides of 3

Here's the way I'd do 3......

1.   angle  XUY  = 180                                                 XUY  is a straight- angle

2.   angle XUW + angle YUW  =                                 angle addtiton theorem

      angle XUY

3.   90  +  angle YUW  = 180                                      substitution, angle XUW = angle 1  = 90

4.   angle YUW = 180 - 90                                           subtract 90 from both sides of 3

5.  angle YUW  =  90  = angle 2                                   simplification  of 4    

Here's the diagram.......

The inradius  will be located at (8,n)   where n is the y coordinate we are looking for....let's call this "O"

The equation of the line containing segment KL  is  y = 0

And the equation of the line that contains segment JK   is  y = (15/8)x →  15x - 8y = 0

And  "O" =  (8,n) will be equidistant from JK and KL

Using the "formula"    for the distance from a point to a line, we have

abs(15(8) - 8n)  / sqrt [ 15^2 + (-8)^2 ]      =  abs ( n )     

And we will assume that both n  and 15(8) - 8n > 0....so we have

[120 - 8n] / 17   = n     

[120 - 8n ] = 17n

120  = 25n

n = 120/25 =  24/5

And the distance ffrom this point to the line y = 0   is just 24/5........and this is the inradius

The equation of the incircle of triangle JKL  is   (x - 8)^2 + (y - 24/5)^2  = (24/5)^2

Thanks, Riddler........5 pts. from me.......!!!!

If you give a good answer to a question, you usually get 5 points from me......other members are able to give you points, too......

But.....don't worry about it too much.......it's just a " fun"   thing....

Mainly.....we're glad you joined, Riddler......Welcome  !!!!

From the associated diagram below, it is obvious that the radius of the circle on which each of the vertices of triangle JKL lie will be located at (8, n)........where n is the y coordinate that we're looking for....let's call this point "O"

We can equate distances here to solve for n

The distance from K to O    =  sqrt [ 8^2 + n^2]   = sqrt [ 64 +n^2 ]

And the distance from  J to O   = sqrt  [ (15 - n)^2]  =  15 - n

So .......sqrt [ 64] + n^2] = 15 - n     ...... square both sides

64 + n^2  =  225 - 30n + n^2

30n  = 225 - 64

30n =  161

n = 161 / 30

So, "O"  =  (8, 161/30)

And the distance from J  to O   =  15 - 161/30   =  [450 -  161] / 30   =  289/30 .....  this is the circumradius......!!!

And the equation of the circle through all three points is :

(x - 8)^2 + (y - 161/30)^2   = (289/  30)^2

P.S.  = There is a "formula" for finding the circumradius of a triangle  =    [ a* b * c ] / [ 4 A]    where a,b,c are the sides of the triangle and A is the area......

Using this, we have   [ 16 * 17^2]  / [ 4 * 120]  = 289 / 30   .....the same result as mine  !!!!

Here's the proof of this "formula"...it's surprisingly simple !!!!........http://www.artofproblemsolving.com/wiki/index.php/Circumradius

Here is what I believe you describe :

AD bisects angle CAB, CM is the perpendicular bisector of AB

XM = 9, AB = 40

And triangle AMX ≈ triangle AEX  .....so..... AX/MX  = AX/EX .......so......MX = EX     ....and  EX is the distance from X to AC = 9

Very nice, Melody!!!!........great diagrams, too....!!!!!

Here's an alternative method.....

Let the side of the square = (1/4)P.....then the diagonal = [ sqrt(2)/4] P    ....and the radius = [sqrt(2)/8] P

Let the side of the equilateral triangle = (1/3)P.....then by the Law of Sines, the radius of  circle  B can be forund thusly :

(1/3)P / sin(120) = r / sin (30)

r =  [(1/2) * (1/3)P] / [sqrt(3)/2] =  [(1/3)P] / sqrt(3)

And, by Euclid, circles are to each other as the squares of their radiuses....so we have....

A / B   =  ( [sqrt(2)/8] P )^2  /  (   [(1/3)P] / sqrt(3) )^2    =   [ 1/32 ]  /  [ 1/27]  =  27/32

11:44

Here's the graph in degrees  [  Pi = 180° ] ..........................https://www.desmos.com/calculator/6agjoftrd9