100. A circle of radius $R$ touches the line $l$ at point $A$, $AB$ is a diameter of this circle, and $BC$ is an arbitrary chord. Let $D$ be the foot of the perpendicular dropped from $C$ to $AB$. Point $E$ lies on

2 I. $\Phi$, Sharygin
the extension of $CD$ beyond point $D$, such that $|ED|=|BC|$. The tangents to the circle passing through $E$ intersect the line at points $K$ and $N$. Find the length of the segment $|KN|$.
100. Let (Fig. 25) $P$ and $Q$ be the points of tangency of the tangents drawn from $E$. We will prove that $|E P|=|E Q|=|B D|$. Indeed,

\[
\begin{aligned}
|E P|^{2}=(|E D| + |D C|) & (|E D| - |D C|)= \\
& =|E D|^{2}-|D C|^{2}=|B C|^{2}-|D C|^{2}=|B D|^{2}
\end{aligned}
\]

(by the condition $|E D|=|B C|$). Let us denote

\[
|K N|=x, \quad|P N|=|N A|=y, \quad|E Q|=|E P|=|B D|=z
\]

Then $|K E|=x+y-z$. We have

\[
S_{K E N}=\frac{1}{2}|K N| \cdot|D A|=\frac{1}{2} x(2 R-z)
\]

on the other hand,

\[
\begin{aligned}
S_{K E N} & =S_{K O N}+S_{E O K}-S_{E O N}= \\
& =\frac{1}{2} R[x+x+y-z-y-z]=R(x-z)
\end{aligned}
\]

(O is the center of the circle). Thus, $\frac{1}{2} x(2 R-z)=R(x-z)$, $x=2 R$.

Answer: $2 R$.