To solve for \( m \angle A \) given the conditions in the problem, we can follow these steps:
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Understand the given information:
- Since \( \overline{BE} \) is perpendicular to \( \overline{ED} \), it means \( \angle EBD = 90^\circ \).
- We have \( \overline{AC} \cong \overline{BC} \), which implies that triangle \( ABC \) is an isosceles triangle with \( AB = AC \).
- It’s given that \( m \angle D = 48^\circ \).
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Using the properties of the triangle:
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In triangle \( BDE \):
- \( m \angle BDE + m \angle D + m \angle EBD = 180^\circ \)
- We have \( m \angle EBD = 90^\circ \) and \( m \angle D = 48^\circ \).
- Thus, we can find \( m \angle BDE \):
\[ m \angle BDE + 48^\circ + 90^\circ = 180^\circ \] \[ m \angle BDE + 138^\circ = 180^\circ \] \[ m \angle BDE = 180^\circ - 138^\circ = 42^\circ \]
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Using the isosceles triangle property:
- In triangle \( ABC \), since \( AC \cong BC \), angles \( A \) and \( C \) are equal:
- Let \( m \angle A = m \angle C = x \).
\[ m \angle A + m \angle C + m \angle B = 180^\circ \] \[ x + x + m \angle B = 180^\circ \] \[ 2x + m \angle B = 180^\circ \]
We need to relate \( m \angle B \) with the angles of triangle \( BDE \).
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Determine \( m \angle B \):
- Angle \( B \) in triangle \( ABC \) is supplementary to angle \( BDE \), since they share side \( \overline{BE} \).
\[ m \angle B = 180^\circ - m \angle BDE = 180^\circ - 42^\circ = 138^\circ \]
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Substituting back: Plugging \( m \angle B \) into the equation for \( m \angle A \):
\[ 2x + 138^\circ = 180^\circ \] \[ 2x = 180^\circ - 138^\circ = 42^\circ \] \[ x = \frac{42^\circ}{2} = 21^\circ \]
Therefore, \( m \angle A \) is
\[ \boxed{21^\circ}. \]