To solve for \( m \angle ADU \) in a cyclic quadrilateral where the vertices are inscribed in a circle, we can use the properties of inscribed angles and opposite angles.
In a cyclic quadrilateral, the opposite angles are supplementary. Therefore, we have:
\[ m \angle AUD + m \angle UAD = 180° \]
We are given:
- \( m \angle UAD = 100° \)
Thus, \[ m \angle AUD = 180° - m \angle UAD = 180° - 100° = 80° \]
Next, considering the angles in the quadrilateral \( A, U, D, Q \):
- \( m \angle AUD = 80° \)
- \( m \angle QU = 55° \)
- \( m \angle AUQ = 112° \)
Now, we need to find \( m \angle ADU \). In the cyclic quadrilateral \( ADQU \), the angles \( \angle AUQ \) and \( \angle ADU \) are also supplementary, since they are opposite angles:
\[ m \angle AUQ + m \angle ADU = 180° \]
Substituting the known value of \( m \angle AUQ \):
\[ 112° + m \angle ADU = 180° \]
Now, solving for \( m \angle ADU \):
\[ m \angle ADU = 180° - 112° = 68° \]
Thus, the measure of \( m \angle ADU \) is:
\[ \boxed{68°} \]
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