Find the 3 radius in the picture below

Question

If ABCD is a square show that: $r1=\frac{l}{16}$, $r2 = \frac{l}{6}$ and $r3 = \frac{3l}{8}$

My progress:

Euclid's Theorem: $\triangle AO2D\\ AO_2^2=DO_2^2+l^2-2r_2l\implies(l+r_2)^2 = (l-r_2)^2+l^2-2r_2l\\ 6lr_2 = l^2 \therefore \boxed{ r_2=\frac{l}{6}}$

$\triangle AJI: (l-r_3)^2=r_3^2+(\frac{l}{2})^2 \implies l^2-2lr_3+r_3^2=r_3^2+\frac{l^2}{4}\\ 4l^2-8lr_3=l^2\rightarrow 3l^2 = 8lr_3\therefore \boxed{r_3=\frac{3l}{8}} $

I can demonstrate $r2$ and $r3$ the Radius $r1$ are missing

Answer 1

For $r_1$ you can use Pythagoras:

$$\left(\frac{l}{2}\right)^2+(l-r_1)^2=(l+r_1)^2$$

This gives the required result straightaway.

Likewise, for $r_3$,

$$r_3^2+\left(\frac{l}{2}\right)^2=(l-r_3)^2$$

Answer 2

Similarly, for $r_3$, use Pythagoras on the triangle whose hypotenuse is the segment from $A$ to the center of that circle and whose third vertex is the point of tangency of that circle with $AD$, giving $$(l-r_3)^2 = r_3^2 + \left(\frac{l}{2}\right)^2.$$