## Task B-4.4.

From point $A$ outside a circle, tangents are drawn to circle $k$ with points of tangency at points $B$ and $C$. A line parallel to $A C$ passes through point $B$ and intersects circle $k$ again at point $D$. If $|A B|=49$ and $|C D|=28$, determine the length of $|A D|$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_aa38f3c63b8013e55583g-20.jpg?height=596&width=882&top_left_y=1935&top_left_x=587)

If tangents are drawn from point A outside the circle, then

$|A C|=|A B|=49$ and $\varangle B C A=\varangle C B A$. Let's denote this angle as $\alpha$.

Then $\varangle B S A=90^{\circ}-\varangle S A B=\alpha$.

Due to the parallelism of lines $A C$ and $B D$, $\varangle C B D=\varangle B C A=\alpha$.

Angle BDC is the smaller inscribed angle over the chord $\overline{B C}$, so it is equal to half of the corresponding central angle, i.e., $\varangle B D C=\varangle B S A=\alpha$.

The same conclusion can be reached by applying the theorem about the angle between a tangent and a chord.

From all the above, it follows that triangle $D B C$ is isosceles, i.e., $|B C|=|D C|=28$.

Then, from half of triangle $B C A$,

$\cos \alpha=\frac{14}{49}=\frac{2}{7}$, and from half of triangle $B C D$,

$\frac{|B D|}{2}=28 \cdot \cos \alpha$, so $|B D|=56 \cdot \frac{2}{7}=16$.

Applying the cosine rule to triangle $A D B$, we have:

$$
\begin{gathered}
|A D|^{2}=|A B|^{2}+|B D|^{2}-2|A B||B D| \cos (2 \alpha) \\
|A D|^{2}=49^{2}+16^{2}-2 \cdot 49 \cdot 16 \cdot\left(2 \cos ^{2} \alpha-1\right)=2401+256-2 \cdot 16 \cdot 49\left(-\frac{41}{49}\right)=3969
\end{gathered}
$$

Finally, the desired distance is $|A D|=\sqrt{3969}=63$.