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 $△BEM\equiv △BCM$, with $\angle BME=\angle BMC=90^{\circ }$. Since $|BC|=|CD|$ and $\angle MBC=90^{\circ }-\angle BCM=\angle FCD$ we get $△FCD\equiv △OBC$. Thus $|FD|=|OC|=\frac{1}{2}|AD|$ and $|DK|=\frac{1}{2}|FD|=\frac{1}{4}|AD|$.

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Alternative solution.

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, $△EMB\equiv △CMB$ 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 $$△KDO\sim △DNO\equiv △ENO\equiv △OME\equiv △OMC\sim △OCB.$$ Thus $\frac{|KD|}{|OD|}=\frac{|OC|}{|CB|}=\frac{1}{2}$ and finally $|AD|=2|OD|=4|KD|$.

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Alternative solution.

We let $\varphi =\angle OBC$. Then $\tan \varphi =\frac{|OC|}{|BC|}=\frac{1}{2}$. The right angled triangles $OBE$ and $BOC$ are congruent, because they share the hypotenuse and $|OE|=|OC|$ (radii of $\Gamma$). Consequently $\varphi =\angle OBC=\angle OBE=\frac{1}{2}\angle EBC$. The double angle formula for tan then gives $$\tan (\angle EBC)=\tan (2\varphi )=\frac{2\tan \varphi }{1-{\tan }^2\varphi }=\frac{4}{3}.$$ 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 $|AD|=4|DK|$.

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Alternative solution.

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 $x=4y$.

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Alternative solution.

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 $x{x}_0+y{y}_0=1$ 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 $$2y_0+x_0=1,x_0^2+y_0^2=1$$ 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 $\frac{1}{2}=\frac{1}{4}|AD|$.