IRL ABooklet

Let $O$ be the midpoint of $CD$, the centre of circle $\Gamma$. Let $F$ be the point where line $CE$ meets $AD$, and let $CE$ meet $OB$ at $M$.

Angle $\angle DEC$ is standing on the diameter $CD$ of $\Gamma$, hence is a right angle. Therefore, triangle $DEF$ has a right angle at $E$. The two tangents $KD$ and $KE$ of $\Gamma$ have the same length: $|KD| = |KE|$. This implies that $K$ is on the perpendicular bisector of $DE$ and so must be the circumcentre of triangle $DEF$. Therefore, $K$ is the midpoint of $DF$.
Because $BC$ and $BE$ are tangents to $\Gamma$, we have $|BE| = |BC|$. Hence $\triangle BEM \equiv \triangle BCM$, with $\angle BME = \angle BMC = 90^\circ$. Since $|BC| = |CD|$ and $\angle MBC = 90^\circ - \angle BCM = \angle FCD$ we get $\triangle FCD \equiv \triangle OBC$. Thus $|FD| = |OC| = \frac{1}{2}|AD|$ and $|DK| = \frac{1}{2}|FD| = \frac{1}{4}|AD|$.

Let $KO$ meet $DE$ at $N$ and let $BO$ meet $EC$ at $M$.
The right angled triangles $BOE$ and $BOC$ are congruent since they share the hypotenuse and $|OE| = |OC|$ (radii). As a consequence, $\triangle EMB \equiv \triangle CMB$ by SAS. This shows that $OB$ is the perpendicular bisector of $EC$. Similarly, $OK$ is the perpendicular bisector of $DE$. Hence $EMON$ is a rectangle and
$\triangle$ KDO $\sim \triangle$ DNO $\equiv \triangle$ ENO $\equiv \triangle$ OME $\equiv \triangle$ OMC $\sim \triangle$ OCB.
Thus $\frac{|KD|}{|OD|} = \frac{|OC|}{|CB|} = \frac{1}{2}$ and finally $|AD| = 2|OD| = 4|KD|$.

We let $\phi = \angle OBC$. Then $\tan \phi = \frac{|OC|}{|BC|} = \frac{1}{2}$. The right angled triangles $OBE$ and $BOC$ are congruent, because they share the hypotenuse and $|OE| = |OC|$ (radii of $\Gamma$). Consequently $\phi = \angle OBC = \angle OBE = \frac{1}{2}\angle EBC$. The double angle formula for tan then gives
$$\tan(\angle EBC) = \tan(2\phi) = \frac{2 \tan \phi}{1 - \tan^2 \phi} = \frac{4}{3}.$$
As $\angle ABK = 90^\circ - \angle EBC$, we obtain
$$\frac{|AK|}{|AB|} = \tan(\angle ABK) = \frac{1}{\tan(\angle EBC)} = \frac{3}{4}.$$
It follows that $|AK| = \frac{3}{4}|AD|$ and so $|AD| = 4|DK|$.

We have $|BC| = |BE|$ (tangents from $B$) and $|KE| = |KD|$ (tangents from $K$). We let $x = |BE| = |BC| = |AB|$ and $y = |KE| = |KD|$.

Triangle $ABK$ has a right angle at $A$ and Pythagoras' Theorem gives
$$(x + y)^2 = |BK|^2 = |AB|^2 + |AK|^2 = x^2 + (x - y)^2.$$
This simplifies to $4xy = x^2$, hence $x = 4y$.

Let the vertices of the square be $A(-1, 2)$, $B(1, 2)$, $C(1, 0)$, $D(-1, 0)$, so that $x^2 + y^2 = 1$ is the circle in question. If $x_0^2 + y_0^2 = 1$, the line $L$ with equation $xx_0 + yy_0 = 1$ is a tangent to the circle at $E(x_0, y_0)$. It passes through $B$ iff $2y_0 + x_0 = 1$. Solving the equations
$$2y_0 + x_0 = 1, \quad x_0^2 + y_0^2 = 1$$
we get $(x_0, y_0) = (-\frac{3}{5}, \frac{4}{5})$ or $(x_0, y_0) = (1, 0)$, and $E$ is the first of these. Hence $L$ has equation $4y - 3x = 5$. This line $L$ meets the line $x = -1$ at $K(-1, \frac{1}{2})$, whose distance from $D(-1, 1)$ is $\frac{1}{2} = \frac{1}{4}|AD|$.