Partitioned Triangle

Claim: $AB=CE+CD$ Proof: Note that the perpendiculars drawn from $I$ to the sides $AB, BC$ and $CA$ all have length $r$ (inradius), since the foot of the perpendiculars (from $I$ ) lie on the incircle

So we can rewrite the given area condition as, $\begin{aligned} [\triangle AIB] &=[\triangle CIE]+[\triangle CID] \\ \\ \frac{AB\cdot r}{2} &=\frac{CE\cdot r}{2}+\frac{CD\cdot r}{2} \\ \\ \implies AB &=CE+CD\end{aligned}$

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Note that the incenter , $I$ lies on the angle bisectors of $\angle A$ and $\angle B$ . Using the angle bisector theorem , $\frac{AE}{CE}=\frac{x}{\sqrt{x^2+16}}$ $\implies CE+CE\cdot\frac{x}{\sqrt{x^2+16}}=4\implies CE=\frac{4\sqrt{x^2+16}}{x+\sqrt{x^2+16}}$ Similarly, $\frac{BD}{CD}=\frac{x}{4}$ $\implies CD+CD\cdot\frac{x}{4}=\sqrt{x^2+16}\implies CD=\frac{4\sqrt{x^2+16}}{x+4}$

Using the above claim, $\begin{aligned} AB&=CE+CD \\ \implies x&=\frac{4\sqrt{x^2+16}}{x+\sqrt{x^2+16}}+\frac{4\sqrt{x^2+16}}{x+4}\end{aligned}$ Solving the above equation gives $x=\pm 2\sqrt{2+2\sqrt{5}}\approx 5.08807859$ $\implies \lfloor10^4\cdot x\rfloor=\boxed{50880}$

Let $BC = y = \sqrt{x^2 + 4^2}$ and the inradius be $r$ . Then the area of $\triangle AIB$ , $[AIB] = \dfrac {rx}2$ and that of quadrilateral $EIDC$ :

$\begin{aligned} [EIDC] & = \blue{\frac {CE}{CA}}[CIA] + \blue{\frac {CD}{BC}}[BIC] & \small \blue{\text{By angle bisector theorem,}} \\ & = \blue{\frac y{x+y}}\cdot \frac {4r}2 + \blue{\frac 4{x+4}}\cdot \frac {ry}2 & \small \blue{\frac {CE}{AE} = \frac {BC}{AB} = \frac yx, \frac {CD}{BD} = \frac 4x} \\ & = \frac {2ry}{x+y} + \frac {2ry}{x+4} & \small \blue{\text{Since }[EIDC]=[AIB]} \\ \frac {rx}2 & = \frac {2ry}{x+y} + \frac {2ry}{x+4} \\ \implies x & = \frac {4y}{x+y} + \frac {4y}{x+4} \\ &= \frac {4\sqrt{x^2+16}}{x + \sqrt{x^2+16}} + \frac {4\sqrt{x^2+16}}{x+4} & \small \blue{\text{Solving the equation}} \\ \implies x & = 2\sqrt{2+2\sqrt 5} \approx 5.088078598 \end{aligned}$

Therefore $\lfloor 10^4 x \rfloor = \boxed{50880}$ .

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Reference: Angle bisector theorem