Easy geometry!

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If we draw a circle with center at B and radius equal to |AB| = 1, we know that the perimeter of any triangle DEF, which has got EF tangent to the circle is equal to 2, because |AE|=|EG| and |FG|=|FC| so the perimeter of triangle DEF is equal to |AD| + |CD| = 2. Now to prove that these triangles DEF are the only ones with perimeter 2 we can consider this: Draw any other line segment E´F´ and we can for every E´F´ find paralel line which is tangent to the quarter circle. Because
E´F´ and the new tangent lines are paralel we can say that triangles DE´F´ and a new triangle made by the paralel line are similar with coeficient of similarity not equal to 1 (because line E´F´ is not same as the new line) so the perimeter is not equal to 2. Now we can easly calculate angle EBF as a half of the sum of angles ABG and GBC so it is 45°.

Let $DF = x$ and $DE=y$ . Since the perimeter of $\triangle DEF$ is 2,

$\begin{aligned} x + y + \sqrt{x^2+y^2} & = 2 \\ x+y - 2 & = \sqrt{x^2+y^2} & \small \color{#3D99F6} \text{Squaring both sides} \\ x^2 + 2xy + y^2 - 4(x+y) + 4 & = x^2+y^2 \\ \implies xy & = 2(x+y - 1) \end{aligned}$

Let $\angle EBF = \theta$ . Then

$\begin{aligned} \theta & = 90^\circ - \tan^{-1} (1-x) - \tan^{-1} (1-y) \\ & = 90^\circ - \tan^{-1} \left(\frac {1-x+1-y}{1-(1-x)(1-y)} \right) \\ & = 90^\circ - \tan^{-1} \left(\frac {2-x-y}{x+y-\color{#3D99F6} xy} \right) & \small \color{#3D99F6} \text{Note that }xy = 2(x+y-1) \\ & = 90^\circ - \tan^{-1} \left(\frac {2-x-y}{x+y-\color{#3D99F6} xy} \right) \\ & = 90^\circ - \tan^{-1} \left(\frac {2-x-y}{2-x-y} \right) \\ & = 90^\circ - \tan^{-1} \left(1 \right) \\ & = 90^\circ - 45^\circ \\ & = = \boxed{45^\circ} \end{aligned}$