To solve this problem, we'll first need to understand some properties of a rhombus and then use the given information about the angles in the rhombus ABCD to find \( m\angle D \).
In a rhombus, the opposite angles are equal, and adjacent angles are supplementary. This means:
- \( m\angle A = m\angle C \)
- \( m\angle B = m\angle D \)
Given that \( m\angle C = 110° \), we can find \( m\angle A \):
\[ m\angle A = 110° \]
Now, since angles A and B are supplementary to angle C and D, respectively:
\[ m\angle A + m\angle B = 180° \] \[ 110° + m\angle B = 180° \] \[ m\angle B = 180° - 110° = 70° \]
Since \( m\angle B = m\angle D \), we have:
\[ m\angle D = 70° \]
Next, since we are interested in showing that diagonals BD and AC are perpendicular using triangle DEC, we note that the angles in triangle DEC must also satisfy the fact that the sum of angles in a triangle is \( 180° \).
Given:
- \( m\angle D = 70° \)
- \( m\angle C = 110° \)
By the CPCTC (Corresponding Parts of Congruent Triangles are Congruent) theorem, if triangle DEC and triangle ABE are created using diagonals, angles C and D in these triangles correspond to triangles ABC and ABD.
Thus, the remaining angle \( m\angle E \) in triangle DEC can be found as:
\[ m\angle E = 180° - m\angle C - m\angle D \] \[ m\angle E = 180° - 110° - 70° = 0° \]
Since we can see from the work above that the diagonals intersect at right angles within the triangle framework being created by the intersection of the diagonals in the rhombus.
Thus, the correct answer for \( m\angle D \) in triangle DEC is:
70°
Because \( m\angle D \) in triangle ABC is equal to \( m\angle B \) as identified earlier, demonstrating the property that confirms the diagonals are perpendicular.
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