Find the radius $r$ of the figure below

Question

Foe reference: Find the radius $r$ of the figure below where $D, F$ and $G$ are points of tangency and $AC = 5$ and $AB = 12$.(S: $r=4$)

My progress:

$\triangle ABC: BC^2 = \sqrt{5^2+12^2} \therefore BC = 13$

H is the center of the larger circle so:

$BH = \frac{13}{2} = 6,5\\ \triangle BIH \sim \triangle BAC \implies \frac{13}{6,5} = \frac{5}{HI} \therefore HI = 2,5\\ \triangle CKF \sim \triangle JKE \implies \frac{5-r}{JE} = \frac{r-KE}{KE}=\frac{r-KJ}{KJ}\\ \triangle BJD \sim\triangle BAC \implies\frac{12}{5} = \frac{12-r}{r-JE} $

???

Answer 1

Consider $EH$ and its "horizontal" and "vertical" components,

$$\begin{align*} (GH-r)^2 &= (AI-r)^2 + (r-HI)^2\\ (6.5-r)^2 &= (6-r)^2 + (r-2.5)^2\\ 6.5^2 - 13r + r^2 &= 6^2 - 12r + r^2 + r^2-5r + 2.5^2\\ 0 &= r^2 - 4r\\ r &= 4 \end{align*}$$

Answer 2

$r = 2\overline{IH}$ = is the side of the square FOAD

Lem de Verriers: $I$ is the incenter of $\triangle ABC$ and is the midpoint of $\overline{FD}$

Inradius $\triangle ABC$:

$2r_{ABC} +a = b+c\implies 2r_{ABC}+13 = 12+5\\ \therefore r_{ABC} =2 \\ \therefore r = 2r_{ABC} = 4$

(By:Palacios)