Determining the True Statement About Angles in a Geometric Diagram
Let angle BCD = x
Since AB \cong DE and \angle A = \angle D (given m \angle B = 43^{\circ}), triangles ABC and CDE are congruent by the ASA (Angle-Side-Angle) criterion.
Therefore, m \angle BCE = m \angle BCD = x (corresponding angles of congruent triangles are equal).
In \triangle BCE,
m \angle BCE + m \angle BCE + m \angle CEF = 180^{\circ} (sum of angles in a triangle)
x + x + 152^{\circ} = 180^{\circ}
2x = 180^{\circ} - 152^{\circ}
2x = 28^{\circ}
x = 14^{\circ}
Since m \angle BCE = x, m \angle BCE = 14^{\circ}. But there is no statement that says m \angle BCE = 14^{\circ}.
Now we need to check angle ACD:
angle ACD = angle BCE (by congruent triangles ABC and CDE)
angle ACD = 14^{\circ}
According to the choices given:
\text{If statement 3 is } \angle ACD = 71^{\circ}, \text{then it is false, as we calculated it to be } 14^{\circ}.
The only statement we did not refute directly is statement 1:
\text{If statement 1 is } m \angle D = 28^{\circ}, \text{ angles D and B would sum to } 43^{\circ} + 28^{\circ} = 71^{\circ}, \text{ which is not the straight angle sum of } 180^{\circ}. \text{ Therefore, statement 1 is false.}
Assuming the diagram and markings are accurate (as we are bound by the image provided), we have shown that statement 1 and statement 3 are both incorrect.
Statement 2 asserts m \angle A = 43^{\circ}, which we can infer from the congruence of \triangle ABC and \triangle CDE since AB \cong DE and \angle B \cong \angle D.
So by process of elimination and confirming with the congruent triangles, the true statement is:
\text{Statement 2: } m \angle A = 43^{\circ}.
