In the diagram of the rhombus, m angle EDC = 42 deg and DE is perpendicular to AC. Using what you know about perpendicular bisectors of a rhombus, what is the m<ECD?

Question

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Answer 1

In a rhombus, the diagonals bisect each other at right angles, and they also bisect the angles at each vertex. Given that \( DE \) is perpendicular to \( AC \), this means that \( DE \) is one of the diagonals, and \( AC \) is the other diagonal.

Since \( DE \) is perpendicular to \( AC \), we know that triangle \( EDC \) is a right triangle, with \( m \angle EDC = 42^\circ \).

In triangle \( EDC \), the angles must add up to \( 90^\circ \) because \( \angle EDC \) is a right angle (\( 90^\circ \)) and \( m \angle ECD \) can be calculated as follows:

\[ m \angle ECD = 90^\circ - m \angle EDC \]

Substituting in the value of \( m \angle EDC \):

\[ m \angle ECD = 90^\circ - 42^\circ = 48^\circ \]

Therefore, the measure of angle \( ECD \) is:

\[ \boxed{48^\circ} \]