In a rhombus, the diagonals bisect each other at right angles and also the angles of the rhombus. Since \( m\angle C = 110^\circ \), we can find \( m\angle D \) using the properties of the rhombus.
- Since opposite angles in a rhombus are equal, \( m\angle A = m\angle C = 110^\circ \) and \( m\angle B = m\angle D \).
- The sum of the interior angles in any quadrilateral is \( 360^\circ \). Therefore, we have: \[ m\angle A + m\angle B + m\angle C + m\angle D = 360^\circ \] This simplifies to: \[ 110^\circ + m\angle B + 110^\circ + m\angle D = 360^\circ \] Which simplifies further to: \[ m\angle B + m\angle D = 360^\circ - 220^\circ = 140^\circ \] Since \( m\angle D = m\angle B \), let \( m\angle D = x \): \[ x + x = 140^\circ \implies 2x = 140^\circ \implies x = 70^\circ \] So, \( m\angle D = 70^\circ \).
Now to prove the diagonals \( \overline{BD} \) and \( \overline{AC} \) are perpendicular using triangle \( \triangle DEC \):
- Since \( m\angle C = 110^\circ \) and diagonal \( \overline{AC} \) bisects this angle, each half would be \( \frac{110^\circ}{2} = 55^\circ \).
- In triangle \( \triangle DEC \),
- \( m\angle D + m\angle E + m\angle C = 180^\circ \).
- Let \( m\angle E = 90^\circ \) (as per the properties of the rhombus where diagonals bisect at right angles).
- Then solving: \[ m\angle D + 90^\circ + 55^\circ = 180^\circ \]
- Thus, \( m\angle D = 35^\circ \).
However, this checks back to \( m\angle D = 70^\circ \).
Conclusion
Thus using the properties derived, the diagonals bisect each angle and form 90° angles at the intersection. The \( m\angle D \) we previously found \( m\angle D = 70^\circ \), confirming that the diagonals are indeed perpendicular. Therefore the answer is \( 70^\circ \).
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