To solve the angles for quadrilateral ABCD, which is a rhombus, we can use the properties of special quadrilaterals.
- Rhombus Properties:
- In a rhombus, opposite angles are equal and adjacent angles are supplementary (they sum to 180°).
Given that \( m∠DAB = 124° \), we can find the following:
- Finding \( m∠ABC \): Since angle DAB and angle ABC are adjacent angles in the rhombus, they are supplementary: \[ m∠ABC = 180° - m∠DAB = 180° - 124° = 56° \]
So, \( m∠ABC = 56° \).
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Finding \( m∠AEB \): Point E is the intersection of the diagonals in rhombus ABCD. The diagonals of a rhombus bisect each other at right angles. Thus, triangle AEB will have angles AEB and ABE as:
We know that \( m∠DAB = 124° \), therefore \( m∠AEB \) can be calculated by recognizing that angle ADB is a straight angle formed by the diagonal. Considering \( m∠DAB = 124° \), \( m∠AEB = \frac{1}{2} \times m∠DAB = \frac{1}{2} \times 124° = 62° \).
So, \( m∠AEB = 62° \).
- Finding \( m∠ADE \): In rhombus ABCD, angle DAB is \( 124° \) and angle DAE and DAB share vertex A. Then angle DAE is half of angle DAB because angle ADE is formed where diagonal AC meets diagonal BD. Hence: \[ m∠ADE = \frac{1}{2} \times (m∠DAB) = \frac{1}{2} \times 124° = 62° \]
Thus, \( m∠ADE = 62° \).
Final Measurements:
- \( m∠ABC = 56° \)
- \( m∠AEB = 62° \)
- \( m∠ADE = 62° \)
These calculations are guided by the properties of a rhombus, specifically the bisection of angles by diagonals and the supplementary nature of adjacent angles.
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