7.41 $D$ is a point on side $CB$ of $\triangle A B C$. If $\angle C A D=\angle D A B=$ $60^{\circ}, A C=3$ and $A B=6$, then the length of $A D$ is
(A) 2 .
(B) 2.5 .
(C) 3 .
(D) 3.5 .
(E) 4 .
(34th American High School Mathematics Examination, 1983)
[Solution 1] Let $A D=y$. Since $A D$ bisects $\angle B A C$, by the angle bisector property of a triangle, we have
$$
\frac{D B}{C D}=\frac{A B}{A C}=2 \text {. }
$$

Therefore, as shown in Figure (1), we set $C D=x, D B=2 x$.
Applying the cosine rule in $\triangle C A D$ and $\triangle D A B$, we get
$$
\left\{\begin{array}{l}
x^{2}=3^{2}+y^{2}-3 y \\
(2 x)^{2}=6^{2}+y^{2}-6 y .
\end{array}\right.
$$
Subtracting (1) multiplied by 4 from (2) gives
$$
0=-3 y^{2}+6 y=-3 y(y-2),
$$

Since $y \neq 0$, we have $y=2$.
Therefore, the answer is $(A)$.
[Solution 2] As shown in Figure (2), extend $C A$ to $E$ such that $B E / / D A$.

Given $\angle A B E=\angle B A D$, and $\angle B E A=\angle D A C$,
we have $\angle E A B=180^{\circ}-120^{\circ}=60^{\circ}$.
Thus, $\triangle A B E$ is an equilateral triangle. Since $\triangle B E C \backsim \triangle D A C$, we have $\frac{D A}{B E}=\frac{C A}{C E}$, or $\frac{D A}{6}=\frac{3}{9}$, so $D A=2$. Therefore, the answer is $(A)$.
[Solution 3] By $S_{\triangle A B C}=S_{\triangle C A D}+S_{\triangle D A B}$, we have
$$
\frac{1}{2} \cdot 3 \cdot 6 \cdot \sin 120^{\circ}=\frac{1}{2} \cdot 3 \cdot A D \cdot \sin 60^{\circ}+\frac{1}{2} \cdot 6 \cdot A D \cdot \sin 60^{\circ} \text {, }
$$

Solving this, we get $A D=2$.
Therefore, the answer is $(A)$.