pls help geometry

From the center M of the circles, draw rays to the corner E and to the center S of the side belonging to the corner. The result is a right-angled triangle M, E, S with sides s/2, r and R.

Then:

\(R^2=(\frac{s}{2})^2+r^2\\r=\sqrt{R^2-(\frac{s}{2})^2}\\ \frac{s}{2}=R\cdot sin(60^{\circ})\)

\(r=\sqrt{R^2-R^2\cdot sin^260^{\circ}}\\ r=R\cdot \sqrt{1-sin^260^{\circ}}\\ r=R\cdot cos\ 60^{\circ}\\ \color{blue}r=\frac{1}{2}R\)

or very simply:

In the right triangle SME, angle SME=60°.

So is:

\(r=R\cdot cos\ 60^{\circ}\\ \color{blue}r=\frac{1}{2}R\)

!

Let A = (0,0)

B= (3,0)

C = (1.5 , 1.5 sqrt 3)  

D  is  the center of  the circumcircle  of ABC   and is  found by the line with the equation

y =   tan (30°) x  =  (1/sqrt3)x

The x coordinate of D  =1.5

So  y = (1/sqrt3) (1.5)  = (3/2)/sqrt3  = sqrt (3) / 2

AD  is the radius of the circumcircle of ABC =  sqrt ( 1.5*2 + (sqrt(3)  / 2)^2 )  = sqrt [ (3/2)^2 + 3/4] = 

sqrt [ 9/4 + 3/4] = sqrt 3

So  the equation of the circumcircle is

(x -1.5)^2  + (y - sqrt (3) / 2)  = 3

F = (1,5 , 0)

E  =  (1.5  ,  sqrt (3)/2 - sqrt (3) )  =  (1.5 , -sqrt (3) / 2 )

The radius of the small circle  = EF / 2 = ( 0 - -sqrt (3)/2 )  /  2 = [ sqrt (3) / 2]  / 2 =   

sqrt (3) / 4

The equation of the small circle is   (x -1.5)^2 + (y - sqrt (3) / 4 )^2   =  3/16