To determine \( m\angle D \) in triangle \( DEC \) using the information provided:
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Rhombus Properties: In a rhombus, the opposite angles are equal and the adjacent angles are supplementary. Since \( m\angle C = 110^\circ \), it follows that \( m\angle A = 110^\circ \) and \( m\angle B = m\angle D = 70^\circ \) (because \( 110^\circ + 70^\circ = 180^\circ \)).
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Triangles Involved: The diagonals of a rhombus bisect its angles. Therefore, diagonals \( AC \) and \( BD \) will create two congruent triangles \( ABD \) and \( CDB \) at point \( E \), where the diagonals intersect.
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Calculating \( m\angle D \): Since \( m\angle C \) is \( 110^\circ \) and the angles \( EDC \) and \( ECD \) are equal (as bisectors) and each measure half of \( m\angle C \), we can say \( m\angle EDC = m\angle ECD = \frac{110^\circ}{2} = 55^\circ \).
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Using Triangle Sum Theorem: In triangle \( DEC \): \[ m\angle D + m\angle E + m\angle C = 180^\circ \] Substituting known angles: \[ m\angle D + 55^\circ + 110^\circ = 180^\circ \] Simplifying gives: \[ m\angle D + 165^\circ = 180^\circ \] \[ m\angle D = 180^\circ - 165^\circ = 15^\circ \]
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Diagonals' Perpendicularity: By the properties of a rhombus, since opposite angles are equal and diagonals bisect the angles, and knowing \( m\angle D = 70^\circ \) (from previous deduction) helps prove that diagonals \( AC \) and \( BD \) are indeed perpendicular.
Since this relates to using CPCTC (corresponding parts of congruent triangles are congruent), and observing the constructed triangles, we can reaffirm that \( m\angle D \) shared here is \( 70^\circ \).
Thus, the correct response would be:
- 70°.
This is correct because it's consistent with the properties of the rhombus, establishing that the diagonals are indeed perpendicular.
3 answers