$\frac{BC}{BA}$

Let $AB = a$ , $BC = b$ and $AC = c$ .

I will not go into the rigor of proving the theorem about in-circles in a right angled triangle, but it is well known that the in-radius, $r$ $=$ $\frac{a + b - c}{2}$

Looking at the bigger circle w.r.t the in-circle, we realize that $r$ $=$ $\frac{EF}{2}$ $\implies$ $EF = a + b - c$

Now considering two right angled triangles, $ABC$ and $AEF$ , we get two equations.

$AC^{2} = AB^{2} + BC^{2}$ $\implies$ $c^{2} = a^{2} + b^{2}$ ... (i)

$AF^{2} = AE^{2} + EF^{2}$ $\implies$ $(\frac{c}{2})^{2}$ $=$ $(\frac{a}{2})^{2}$ $+$ $(a + b - c)^{2}$ ... (ii)

Using (i) and (ii), we get: $c$ $=$ $a$ $+$ $\frac{b}{2}$ ... (iii)

Substituting (iii) back in (i), we get: $\frac{b}{a}$ $=$ $\frac{BC}{BA}$ $=$ $\frac{4}{3}$

Hence, the solution sought $=$ $4 \times 3$ $=$ $\boxed{12}$

You do not have to scratch your head or go through elaborate theorems or extensive calculations, if you know the in-radius relationship.

From the image provided, let $|IP|\perp |AF|$

Let $|FP|=p, |AF|=2R$ , then $|PF|=2R-p$

$AF$ is the diameter, so $\angle AIF=90^{\circ}$

$\Delta AIP\sim \Delta IFP$

Let $|IP|=r$ ,

then, $p(2R-p)=r^{2}$

$E,F$ are midpoints of $AB,AC$ respectively.

$AC=4R$

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Let midpoint of $AF$ be M

Let $MQ\perp AE$

Let $|AQ|=x$

$|AB|=4x$

$|MQ|=\sqrt{R^{2}-x^{2}}$

$|BC|=4 \sqrt{R^{2}-x^{2}}$

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Let $IX\perp AB$

$|AX|=p$

$|BX|=4x-p$

$|BX|=r$

$4x-p=r$

Similiarly, we have $4 \sqrt{R^{2}-x^{2}}-r=4R-p$

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$(1):p(2R-p)=r^{2}$

$(2):4x-p=r$

$(3):4 \sqrt{R^{2}-x^{2}}-r=4R-p$

From $(2)$ , $p=4x-r$

$(3):4 \sqrt{R^{2}-x^{2}}-r=4R-4x+r$

$4 \sqrt{R^{2}-x^{2}}-2r=4R-4x$

$2 \sqrt{R^{2}-x^{2}}-r=2(R-x)$

$2 \sqrt{R^{2}-x^{2}}-2(R-x)=r$

$r=2 \sqrt{R-x}(\sqrt{R+x}-\sqrt{R-x})$

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$(2):p=4x-r$

$p=4x-2 \sqrt{R-x}(\sqrt{R+x}-\sqrt{R-x})$

$p=4x-2 \sqrt{R^{2}-x^{2}}+2(R-x)$

$p=2R+2x-2 \sqrt{R^{2}-x^{2}}$

$p=2 \sqrt{R+x}(\sqrt{R+X}-\sqrt{R-x})$

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$(1):p(2R-p)=r^{2}$

$2Rp-p^{2}=r^{2}$

$2R\cdot 2 \sqrt{R+x}(\sqrt{R+X}-\sqrt{R-x})-[2 \sqrt{R+x}(\sqrt{R+X}-\sqrt{R-x})]^{2}={[2 \sqrt{R-x}(\sqrt{R+x}-\sqrt{R-x}})]^{2}$

$4R\cdot \sqrt{R+x}(\sqrt{R+X}-\sqrt{R-x})-4(R+x)(\sqrt{R+X}-\sqrt{R-x})^{2}=4(R-x)(\sqrt{R+x}-\sqrt{R-x})^{2}$

$4R\cdot \sqrt{R+x}(\sqrt{R+X}-\sqrt{R-x})=4(R+x+R-x)(\sqrt{R+x}-\sqrt{R-x})^{2}$

$4R\cdot \sqrt{R+x}(\sqrt{R+X}-\sqrt{R-x})=4(2R)(\sqrt{R+x}-\sqrt{R-x})^{2}$

$\sqrt{R+x}(\sqrt{R+X}-\sqrt{R-x})=2(\sqrt{R+x}-\sqrt{R-x})^{2}$

$\sqrt{R+x}=2(\sqrt{R+x}-\sqrt{R-x})$

$\sqrt{R+x}=2 \sqrt{R+x}-2 \sqrt{R-x})$

$\sqrt{R+x}=2 \sqrt{R-x}$

$R+x=4(R-x)$

$3R=5x$

$\cfrac{x}{R}=\cfrac{3}{5}$

$\cfrac{|AQ|}{|AM|}=\cfrac{3}{5}$

$\cfrac{|AB|}{|AC|}=\cfrac{3}{5}$

$\cfrac{|BC|}{|BA|}=\cfrac{4}{3}=\cfrac{a}{b}$

$ab=4\times 3=12$

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It is very tricky to juggle with three variables $R, r, x$ at the beginning and $\sqrt{R+x}$ and $\sqrt{R-x}$ at the end. It appeals so much to me how algebraic this geometry problem is!

Let $\omega_1$ be the circumcircle of $\triangle AEF$ and $\omega_2$ be the circumcircle of $\triangle ABC.$ It is clear that there exists a homothety that maps $\triangle AEF$ to $\triangle ABC$ and $\omega_1$ to $\omega_2$ with dilation ratio 2. Letting $X$ be the intersection of line $AI$ with $\omega_2,$ we have $AX = 2AI$ by the homothety, so $AI = IX.$

Let $a = BC, b = AC, c = AB,$ and $n = IX.$ It is well-known that $IX = BX = CX,$ so $BX = CX = n$ and $AX = 2n.$ Applying Ptolemy's Theorem on $ABXC$ yields

$\begin{aligned} AB(CX) + AC(BX) &= BC(AX) \\ cn + bn &= a(2n) \\ c + b &= 2a. \quad \quad \textcolor{cyan}{\small{(*)}} \end{aligned}$

We also have $a^2 + c^2 = b^2$ by the Pythagorean Theorem, and we wish to find $\dfrac{a}{c}.$ From equation $\textcolor{cyan}{\small{(*)}},$ we get $b = 2a - c.$ Substituting this into our Pythagorean Theorem equation yields

$\begin{aligned} a^2 + c^2 &= (2a - c)^2 \\ a^2 + c^2 &= 4a^2 - 4ac + c^2 \\ 3a^2 - 4ac &= 0 \\ a(3a - 4c) &= 0. \end{aligned}$

Since $a \neq 0,$ we must have $3a - 4c = 0,$ or $\dfrac{a}{c} = \dfrac{4}{3}.$ Our final answer is $4(3) = \boxed{12}.$

i just used coordinates. if we take B to be the origins, then the in-centre has coordinates $(r,r)$ with $r= \dfrac{x+y-\sqrt{x^2+y^2}}{2}$ where x,y are side lengths. the circumcentre of the smaller triangle has coordinates $(3/4x,1/4y)$ with circumradius $R= \dfrac{\sqrt{x^2+y^2}}{4}$ . we can now use the equation of a circle to get $(3/4 x-r)^2+(1/4y-r)^2= R^2=\dfrac{x^2+y^2}{16} \to x^2-3xr-yr+4r^2=0 \to 1-3\dfrac{r}{x}-\dfrac{y}{x}\dfrac{r}{x}+4(\dfrac{r}{x})^2=0$ if we define $\dfrac{y}{x}=a$ , then $\dfrac{r}{x}= \dfrac{1+a-\sqrt{1+a^2}}{2}$ . plugging this in and expanding everything: $3 + 3 a^2 -\sqrt{1 + a^2} - 3 a\sqrt{1 + a^2} = 0\to 3(1+a^2)=(3a+1)\sqrt{a^2+1}\to 9(1+a^2)=(3a+1)^2\to 9=6a+1\to a= \boxed{\dfrac{4}{3}}$

You hardly need the circumcircle except to note it lies on the perpendicular bisector of $EF$ and contains $I$ . Let the points of tangency of the incircle be as shown and connected to the incenter.

$BG = \frac{1}{2} EF = \frac{1}{4} BC$ . For simplicity let $BG=1, BC=4, GC=3$ .

Let $BA=x$ so $AC=\sqrt{16+x^{2}}$ . and $AH=x-1$ .

Then by congruent triangles $BH=1$ , $AH=AJ=x-1$ , $CG=CJ=3$ which gives another expression for $AC: x+2$

Solving $16+x^{2}=(x+2)^{2}$ gives $x=3$ so $\frac{BC}{BA}=\frac{4}{3}$ and the solution is then $4 \cdot 3 = \boxed{12}$