4. Among the following 3 geometric propositions,
(1) Two similar triangles, if their perimeters are equal, then these two triangles are congruent;
(2) Two similar triangles, if two sets of sides are equal, then these two triangles are congruent;
(3) Two similar triangles, the angles (not obtuse) formed by their corresponding sides (or their extensions) are equal. The number of true propositions is ( ).
(A) 3
(B) 2
(C) 1
(D) 0
4. (B).

The first proposition is true. Let the side lengths of two similar triangles be $a, b, c$ and $a_{1}, b_{1}, c_{1}$, respectively. Then, by similarity,
$$
\frac{a}{a_{1}}=\frac{b}{b_{1}}=\frac{c}{c_{1}}=\frac{a+b+c}{a_{1}+b_{1}+c_{1}}=1,
$$

which implies $a=a_{1}, b=b_{1}, c=c_{1}$.
Therefore, the two triangles are congruent.
The second proposition is false. When the side lengths of two similar triangles are $8, 12, 18$ and $12, 18, 27$, respectively, they are not congruent.
The third proposition is true. Proof as follows:
In $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$, the corresponding sides $B A$ and $B_{1} A_{1}$ intersect at $D$, and $C A$ and $C_{1} A_{1}$ intersect at $E$. It is sufficient to prove that
$$
\angle A D A_{1}=\angle A E A_{1} \text {. }
$$

Draw $D F / / A C$ and $D G / / C_{1}$, then
$$
\begin{array}{l}
\angle A D \bar{F}=\angle B A C, \\
\angle A_{1} D G=\angle B_{1} A_{1} C_{1}, \\
\angle F D G=\angle A E A_{1} .
\end{array}
$$

Since $\triangle A B C \sim \triangle A_{1} B_{1} C_{1}$, we have
$$
\angle B A C=\angle B_{1} A_{1} C_{1} \Rightarrow \angle A D F=\angle A_{1} D G \text {, }
$$

then
$$
\begin{array}{l}
\angle A D A_{1}=\angle A D F-\angle A_{1} D F \\
=\angle A_{1} D G-\angle A_{1} D F \\
=\angle F D G=\angle A E A_{1}
\end{array}
$$