B4 In the following diagram, two lines that meet at a point $A$ are tangent to a circle at points $B$ and $C$. The line parallel to $A C$ passing through $B$ meets the circle again at $D$. Join the segments $C D$ and $A D$. Suppose $A B=49$ and $C D=28$. Determine the length of $A D$.
Solution 1: The answer is $A D=63$.

Join the segment $B C$. Since the two lines are both tangent to the circle, $A B=A C$. Therefore, $\angle A B C=\angle A C B$.

Furthermore, since $B D$ is parallel to $A C, \angle A C B=\angle D B C$. Since $A C$ is tangent to the circle at $C$, by the tangent-chord theorem, $\angle B D C=\angle A C B$. Hence, we have the following sequence of equal angles:
$$
\angle A B C=\angle A C B=\angle C B D=\angle C D B .
$$

Furthermore, $A B=A C$ and $C B=C D$. Therefore, $\triangle A B C$ is similar to $\triangle C B D$. Hence.
$$
\frac{A B}{B C}=\frac{C B}{B D} \text {. }
$$

Since $A B=49$ and $B C=C D=28, B D=B C^{2} / A B=28^{2} / 49=4^{2}=16$.

Let $M$ be the foot of the perpendicular from $D$ on $A C$ and $N$ the foot of the perpendicular on $B D$ from $C$.

Since $C B=C D, N$ is the midpoint of $B D$. Since $B D$ is parallel to $C M, N D M C$ is a rectangle. Therefore, $C M=N D=\frac{1}{2} \cdot B D=8$. We now determine the length of $D M$.
$$
D M=N C=\sqrt{D C^{2}-D N^{2}}=\sqrt{28^{2}-8^{2}}=\sqrt{784-64}=\sqrt{720} .
$$

Therefore,
$$
A D=\sqrt{A M^{2}+M D^{2}}=\sqrt{(A C+C M)^{2}+M D^{2}}=\sqrt{(49+8)^{2}+720}=\sqrt{3969}=63 .
$$

Hence, the answer is 63 .

Solution 2: As in Solution 1, join segment $B C$ and conclude that $B C=28$ and $B D=16$. Also as in Solution $1, \angle A B C=\angle A C B=\angle C B D=\angle C D B$. Let $\theta$ be this angle.

Let $N$ be the foot of the perpendicular on $B D$ from $C$. As in Solution $1, N$ is the midpoint of $B D$. Therefore, $B N=8$. We can now determine $\cos \theta$ from $\triangle C B N$, which is
$$
\cos \theta=\frac{B N}{B C}=\frac{8}{28}=\frac{2}{7} .
$$
(Alternatively, we can use $\triangle A B C$ to determine $\cos \theta$.) Note that $\angle A B D=2 \theta$. We then apply the cosine law on $\triangle A B D$ to determine $A D$. By the cosine law, we have
$$
A D^{2}=B A^{2}+B D^{2}-B A \cdot B D \cdot \cos (2 \theta) .
$$

We first determine $\cos (2 \theta)$, which by the double-angle formula is
$$
\cos (2 \theta)=2 \cos ^{2} \theta-1=2 \cdot(2 / 7)^{2}-1=-41 / 49 .
$$

Therefore,
$$
A D^{2}=49^{2}+16^{2}-2(49)(16)(-41 / 49)=2401+256+2 \cdot 16 \cdot 41=3969 .
$$

Hence, $A D=\sqrt{3969}=63$.

Solution 3: Let $\theta$ be defined as in Solution 2. Then as shown in Solution $2, \cos \theta=2 / 7$.
Then note that
$$
\angle B C D=180-\angle C B D-\angle C D B=180-2 \theta .
$$

Therefore, $\angle A C D=\angle A C B+\angle B C D=\theta+(180-2 \theta)=180-\theta$. We now apply the cosine law on $\triangle A C D$.
$$
\begin{array}{c}
A D^{2}=C A^{2}+C D^{2}-2 \cdot C A \cdot C D \cdot \cos \angle A C D=49^{2}+28^{2}-2 \cdot 49 \cdot 28 \cdot \cos (180-\theta) \\
=2401+784+2 \cdot 49 \cdot 28 \cdot \cos \theta=3185+2 \cdot 49 \cdot 28 \cdot \frac{2}{7}=3185+4 \cdot 7 \cdot 28=3969 .
\end{array}
$$

Therefore, $A D=\sqrt{3969}=63$.