$17 \cdot 31$ If $\triangle A_{1} A_{2} A_{3}$ is an equilateral triangle, and $A_{n+3}$ is the midpoint of segment $A_{n} A_{n+1}$ (where $n$ is any positive integer), then the degree measure of $\angle A_{44} A_{45} A_{43}$ is
(A) $30^{\circ}$.
(B) $45^{\circ}$.
(C) $60^{\circ}$.
(D) $90^{\circ}$.
(E) $120^{\circ}$.
[Solution] The internal angles of $\triangle A_{2} A_{3} A_{4}$ are $60^{\circ}, 30^{\circ}, 90^{\circ}$.
Since $\angle A_{1} A_{2} A_{3}=60^{\circ}, A_{2} A_{4}=A_{2} A_{5}$,
$\therefore \triangle A_{2} A_{4} A_{5}$ is an equilateral triangle.
Therefore, the vertex angles of $\triangle A_{3} A_{4} A_{5}$ are $30^{\circ}, 30^{\circ}$,
$120^{\circ}$. Then the vertex angles of $\triangle A_{4} A_{5} A_{6}$ are $30^{\circ}, 60^{\circ}, 90^{\circ}$.
Since $\angle A_{4} A_{5} A_{6}=60^{\circ}, A_{5} A_{6}=A_{5} A_{7}$,
$\therefore \triangle A_{5} A_{6} A_{7}$ is also an equilateral triangle. Recursively, we can deduce that $A_{n}$ and $A_{n+4}$ are corresponding angles, and $\triangle A_{n} A_{n+1} A_{n+2} \smile \triangle A_{n+4} A_{n+5} A_{n+6}$.
$$
\therefore \angle A_{44} A_{45} A_{43}=\angle A_{4} A_{5} A_{3}=120^{\circ} \text {. }
$$

Therefore, the answer is $(E)$.