The problem asks us to confirm, using the drawings, that $\angle 1 = \angle 2$ under the given conditions:

1. $AB = BC$ (the segments $AB$ and $BC$ are equal).
2. Point $B$ bisects the straight line $AC$.

Solution Explanation:

For Diagram (a):

1. Since $AB = BC$, $\triangle ABC$ is isosceles with $AB = BC$.
2. In an isosceles triangle, the angles opposite the equal sides are equal. This means: $\angle 1 = \angle 2$
3. The equality $\angle 1 = \angle 2$ is verified.

For Diagram (b):

1. Similarly, since $AB = BC$, $\triangle ABC$ is isosceles with $AB = BC$.
2. Angle $\angle 1$ and $\angle 2$ are base angles of the isosceles triangle $\triangle ABC$, making them equal: $\angle 1 = \angle 2$

Thus, in both cases, the condition $\angle 1 = \angle 2$ holds due to the isosceles triangle property.

Would you like a detailed proof for isosceles triangles or further explanation of the geometry? Here are some follow-up questions to deepen understanding:

1. What are the key properties of isosceles triangles used in this problem?
2. How can we mathematically prove that base angles in isosceles triangles are equal?
3. Why does the condition $AB = BC$ directly lead to the equality of angles?
4. Can we generalize this property for other shapes with symmetry (e.g., rhombuses)?
5. How do line bisectors relate to this kind of geometric reasoning?

Tip: Always identify equal sides or angles in a geometric figure to spot useful properties, like symmetry or congruence.