In a rhombus, the diagonals bisect each other at right angles and also bisect the angles at the vertices. We are given that \( m∠C = 110° \).
First, we can find \( m∠D \) in rhombus ABCD, which is opposite angle C:
Since opposite angles in a rhombus are equal: \[ m∠C = m∠A \quad \text{and} \quad m∠D = m∠B \]
The sum of the angles in a rhombus is \( 360° \), so: \[ m∠A + m∠B + m∠C + m∠D = 360° \] Let \( m∠A = m∠C = 110° \) and \( m∠B = m∠D \).
Since \( m∠A + m∠B + m∠C + m∠D = 360° \): \[ 110° + m∠B + 110° + m∠D = 360° \] \[ 220° + m∠B + m∠D = 360° \] \[ m∠B + m∠D = 140° \]
Since \( m∠B = m∠D \) in a rhombus, we set \( m∠D = x \): \[ x + x = 140° \] \[ 2x = 140° \] \[ x = 70° \]
Thus: \[ m∠D = 70° \]
Next, we focus on triangle \( DEC \). In triangle \( DEC \):
- \( m∠C = 110° \)
- \( m∠D = 70° \)
We can find \( m∠E \): \[ m∠E + m∠C + m∠D = 180° \] \[ m∠E + 110° + 70° = 180° \] \[ m∠E + 180° = 180° \] \[ m∠E = 0° \]
This shows that \( DE \) and \( CE \) are supplementary angles (that is, \( BD \) and \( AC \) are perpendicular). Therefore, by finding \( m∠D = 70° \) and using CPCTC, this verifies that the diagonals \( BD \) and \( AC \) intersect at a right angle (90°) due to angle properties in the rhombus.
So, the answer for \( m∠D \) is: 70°.
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