Strange quadrilateral

Extend $CD$ and $BE$ to meet at $H$ . Then we note circle $k$ is the incircle of $\triangle HBC$ with inradius $r=FG=FI=FJ$ . We note that $\triangle ABE$ and $\triangle HBC$ are similar. Then if $\angle ABE = \theta$ , then $\angle IHF = \dfrac \theta 2$ .

$\begin{aligned} \tan \angle IHF & = \frac {IF}{HI} \\ & = \frac {r}{{\color{#3D99F6}HD}+CD-\color{#D61F06}CI} & \small \color{#3D99F6} \text{Note that }\triangle HDE \text{ and }\triangle ABE \text{ are congruent.} \\ & = \frac {r}{{\color{#3D99F6}AB}+1-\color{#D61F06}FJ} \\ & = \frac {r}{{\color{#3D99F6}1}+1-\color{#D61F06}r} \\ \implies \tan \frac \theta 2 & = \frac r{2-r} \end{aligned}$

We also note that $\tan \theta = \tan \angle ABE = \dfrac {AE}{AB} = \dfrac 12$ , then we have:

$\begin{aligned} \tan \theta & = \frac 12 \\ \frac {2\tan \frac \theta 2}{1-\tan^2 \frac \theta 2} & = \frac 12 \\ \tan^2 \frac \theta 2 + 4 \tan \frac \theta 2 - 1 & = 0 & \small \color{#3D99F6} \text{Putting }\tan \frac \theta 2 = \frac r{2-r} \\ \left(\frac r{2-r}\right)^2 + \frac {4r}{2-r} - 1 & = 0 \\ r^2 - 3r +1 & = 0 \\ \implies r & = \frac {3-\sqrt 5}2 \end{aligned}$

Note that quadrilateral $EGFD$ is made up of $\triangle DEF$ and $\triangle EFG$ , then:

$\begin{aligned} [EGFD] & = [DEF]+[EFG] \\ & = \frac {DI\times DE}2 + \frac {FG\times \color{#3D99F6} EG}2 \\ & = \frac {(1-r)\times \frac 12}2 + \frac {r {\color{#3D99F6}\left(2-\sqrt 5/2 - r\right)}}2 & \small \color{#3D99F6} \text{See note.} \\ & = \frac {1+(3-\sqrt 5)r - 2r^2}4 & \small \color{#3D99F6} \text{Note that } r = \frac {3-\sqrt 5}2 \\ & = \frac {1+2r^2-2r^2}4 \\ & = \frac 14 = \boxed{0.25} \end{aligned}$

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Note:

$\begin{aligned} EG & = {\color{#3D99F6}HG} - {\color{#D61F06}HE} & \small \color{#3D99F6} \text{Note that }\triangle HFG \text{ and }\triangle HFI \text{ are congruent.} \\ & = {\color{#3D99F6}HI} - {\color{#D61F06}BE} & \small \color{#D61F06} \triangle HDE \text{ and }\triangle ABE \text{ are congruent.} \\ & = {\color{#D61F06}HD} + DI - {\color{#D61F06}\frac {\sqrt 5}2} \\ & = {\color{#D61F06}AB} +1-r - \frac {\sqrt 5}2 \\ & = {\color{#D61F06}1} +1-r - \frac {\sqrt 5}2 \\ & = 2 - \frac {\sqrt 5}2 - r \end{aligned}$