Problem Breakdown:

1. $ABCD$ is a rectangle, implying:

   • $AB \perp BC$, $AD \perp AB$, etc.
   • Opposite sides are equal: $AB = CD$ and $BC = AD$.

2. $E$ is the midpoint of $AC$, meaning:

   • $AC$ is a diagonal, and $E$ divides it into two equal halves.

3. $\angle BCA = 58^\circ$, and you are tasked to find $\angle BAC$.

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Solution:

Since $ABCD$ is a rectangle:

• Diagonals $AC$ and $BD$ bisect each other at $E$.
• The diagonals split $\triangle ABC$ into two isosceles triangles: $\triangle ABC$ (isosceles, $AB = BC$).

In $\triangle ABC$:

The sum of angles in a triangle is $180^\circ$. Let $\angle BAC = x$. Then: $x + 58^\circ + \angle ABC = 180^\circ$

From the rectangle property:

• $\angle ABC = 90^\circ$.

Substitute: $x + 58^\circ + 90^\circ = 180^\circ$

Simplify: $x = 180^\circ - 148^\circ = 32^\circ$

Final Answer:

$\boxed{\angle BAC = 32^\circ}$

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Do you want further details or clarification? Here are some related questions to explore:

1. What are the properties of the diagonals in a rectangle, and how do they influence angle calculations?
2. Can you prove that the triangles formed by a diagonal of a rectangle are isosceles?
3. How does knowing $\angle BCA$ help in finding $\angle BAC$ in different quadrilaterals (e.g., parallelogram, square)?
4. What is the relationship between the diagonals of a rectangle and their intersection point?
5. How do changes in $\angle BCA$ affect $\angle BAC$ for other rectangle configurations?

Tip: In problems involving rectangles, always use the right-angle property and diagonal symmetry to simplify calculations!