106. The base of the triangular pyramid $A B C D$ is the triangle $A B C$, in which $\widehat{A}=\pi / 2, \widehat{C}=\pi / 6$, $|B C|=2 \sqrt{2}$. The lengths of the edges $A D, B D$ and $C D$ are equal to each other. A sphere of radius 1 touches the edges $A D, B D$, the extension of the edge $C D$ beyond point $D$, and the plane $A B C$. Find the length of the tangent segment drawn from point $A$ to the sphere.
106. Let $K$ be the point of tangency of the sphere with the extension of $CD$, and $M$ and $L$ be the points of tangency with the edges $AD$ and $BD$, and $N$ be the midpoint of $BC$ (Fig. 14). Since $|CD| = |DB| = |DA|$, $DN$ is perpendicular to the plane $ABC$, $|DK| = |DM| = |DL|$, $KL$ is parallel to $DN$, $ML$ is parallel to $AB$, hence, the plane $KLM$ is perpendicular to the plane $ABC$, $\widehat{KLM} = 90^\circ$. If $O$ is the center of the sphere, then the line $DO$ is perpendicular to the plane $KLM$, i.e., $DO$ is parallel to the plane $ABC$, thus, $|DN| = 1$ (the radius of the sphere). Moreover, $DO$ passes through the center of the circle circumscribed around $\triangle KLM$ - the midpoint of $KM$. From this, it follows that $\widehat{ODM} = \frac{1}{2} \widehat{KDM}$. Further, $|DA| = |DC| = \sqrt{|CN|^2 + |DN|^2} = \sqrt{3}$, $|CA| = |CB| \cos 30^\circ = \sqrt{6}$, i.e., $\triangle CDA$ is a right triangle, $\widehat{CDA} = 90^\circ$, $\widehat{ODM} = 45^\circ$, $|DM| = |OM| = 1$. The desired tangent segment is $|AM| = |AD| - |DM| = \sqrt{3} - 1$.