$17 \cdot 38$ Given a triangle $\triangle ABC$ with acute angles, construct equilateral triangles on each side of $\triangle ABC$ outside the shape, resulting in $\triangle ABD, \triangle BCE, \triangle CAF$. The conclusions are:
(1) $AE = BF = CD$.
(2) $AE, BF, CD$ intersect at one point.
(3) The angle (acute) between any two of $AE, BF, CD$ is $60^{\circ}$.

Then
(A) None of (1)(2)(3) are true.
(B) All of (1)(2)(3) are true.
(C) (1)(3) are true, (2) is false.
(D) (1) is true, (2)(3) are false.
(Shenyang City, Liaoning Province, "Yucai Cup" Junior High School Mathematics Competition, 1988)
[Solution] As shown in the figure, (1) It is easy to prove that $\triangle A B F \cong \triangle A D C, \triangle B D C \cong \triangle B A E$,
thus $B F=C D=A E$.
(2) Assume that $B F$ and $C D$ intersect at point $O$, and connect $O A$ and $O E$.
It is easy to prove that $\triangle A B F \cong \triangle A D C$, so the distance from $A$ to $B F$ and $C D$ is equal, which means $O A$ is the angle bisector of $\angle D O F$.
From $\triangle A B F \cong \triangle A D C$, we know that $\angle A F O=\angle A C O$.
Therefore, $A, F, C, O$ are concyclic, thus $\angle C O F=\angle C A F=60^{\circ}$.
According to the problem, $\angle C E B=60^{\circ}=\angle C O F$, so quadrilateral $O B E C$ has a circumcircle.
Since $E B=E C, \therefore \angle B O E=\angle C O E$, which means $O E$ is the angle bisector of $\angle B O C$. The angle bisectors of vertical angles lie on the same line, so $A, O, E$ are collinear. This means that $A E, B F, C D$ intersect at one point $O$.
(3) From the proof in (2), we know
$$
\angle A O F=\angle A O D=\angle B O D=\angle B O E=\angle C O E=\angle C O F=60^{\circ} \text {. }
$$

Therefore, (1), (2), and (3) are all true.
Hence, the answer is $(B)$.