I wonder what $A$ might be

Let $AD=DB=1$ . Then $DE=\tan A$ , $BC= 2\tan A$ , and the radius of the circle $r = \frac 12$ . We note that

$\begin{aligned} r + \cot \left(\frac {90^\circ - A}2 \right) & = BC \\ \frac 12 + \frac 12 \cot \left(45^\circ - \frac \theta 2 \right) & = 2\tan A & \small \blue{\text{Let }t = \tan \frac A2} \\ \frac 12 + \frac 12 \cdot \frac {1+t}{1-t} & = \frac {2 \cdot 2t}{1-t^2} \\ \frac 12 \cdot \frac 2{1-t} & = \frac {4t}{(1+t)(1-t)} & ...(1) \\ 1+t & = 4t \\ t & = \frac 13 \\ \implies \tan A & = \frac {2t}{1-t^2} & \small \blue{\text{Note (1): }\frac 1{1-t} = \frac {4t}{1-t^2}} \\\ & = \frac 1{2(1-t)} = \frac 34 = \boxed{0.75} \end{aligned}$

Let $AB = 1$ and $BC = x$ , so that we are finding $\tan A = \frac{BC}{AB} = x$ .

Then by the Pythagorean Theorem on $\triangle ABC$ , $AC = \sqrt{x^2 + 1}$ .

Since $AD = DB$ and $AB = 1$ , $AD = DB = \frac{1}{2}$ , and since the radius of the circle is half the height of trapezoid $DBCE$ , $r = \frac{1}{4}$ .

The radius $r$ of the incircle of $\triangle ABC$ is also $r = \frac{1}{2}(x + 1 - \sqrt{x^2 + 1})$ .

Therefore, $\frac{1}{2}(x + 1 - \sqrt{x^2 + 1}) = \frac{1}{4}$ , and this solves to $x = \boxed{0.75}$ .

Let the radius of the circle be $r=1$ and $F$ , $G$ the contact points of $DE$ and $EC$ with the circle, as seen in the figure. Denote $FE$ by $x$ . Then, $DE=1+x$ , $DB=2$ .

Since $DE\parallel BC$ and $D$ is the midpoint of $AB$ , it holds that $BC=2DE=2+2x$ .

From $E$ we drop $EK\bot BC$ .

Then, $EK=DB=2$ , $LK=FE=x$ , $BL=DF=1$ and $KC=BC-BK=1+x$ . $\ \$

Furthermore, $EG=EF=x \ \ \ \ \ \text{(tangent segments)}$ and $GC=LC=LK+KC=x+(1+x)=1+2x$ Thus, $EC=EG+GC=x+(1+2x)=1+3x$

By Pythagora’s theorem on right $\triangle EKC$ , $E{{C}^{2}}=E{{K}^{2}}+K{{C}^{2}}\Leftrightarrow {{\left( 1+3x \right)}^{2}}={{2}^{2}}+{{\left( 1+x \right)}^{2}}\Leftrightarrow 2{{x}^{2}}+x-1=0\overset{x>0}{\mathop{\Leftrightarrow }}\,x=\frac{1}{2}$ Consequently, $\tan A=\tan \left( \angle CEK \right)=\dfrac{KC}{EK}=\dfrac{1+\frac{1}{2}}{2}=\boxed{0.75}$

Let the circle's radius be "r", for generality. Let DE = x. Then BC = 2x (midpoint theorem). By inspection, PE = x-r and EQ = PE (equal tangents). Similarly, RC = 2x-r = CQ. Thus EC = EQ + QC = (x-r) + (2x-r) = 3x-2r. By the Pythagorean theorem, AE = $\sqrt{(2r)^2+x^2}$ . Since AE = EC, $\begin{aligned} \\ \sqrt{(2r)^2+x^2} & =3x-2r \\ 4r^2+x^2 & =9x^2-12xr+4r^2 \qquad \text{(squaring both sides)} \\ -8x^2+12xr & =0 \qquad \text{(simplifying)} \\ 2x-3r & =0 \qquad \text{(dividing by -4x since } x\ne 0) \\ x & =\frac32 r \\ \end{aligned}$ $\therefore \tan A = \frac{DE}{AD} = \frac{\frac32 r}{2r} = \frac34 = \boxed{0.75}$