Question

given:
parallelogram abcd
prove:
consecutive interior angles are
supplementary.
(m\angle b + m\angle a = 180^{circ}
and
m\angle c + m\angle b = 180^{circ})
(draw an auxiliary segment bd)

Explanation:

Step1: Properties of parallelogram

In parallelogram \(ABCD\), \(AB\parallel CD\) and \(AD\parallel BC\) (Definition of parallelogram).

Step2: Alternate - interior angles

Since \(AB\parallel CD\) and \(AD\) is a transversal, \(\angle A+\angle D = 180^{\circ}\) (Consecutive - interior angles theorem: If two parallel lines are cut by a transversal, then consecutive - interior angles are supplementary). Also, since \(AD\parallel BC\) and \(AB\) is a transversal, \(\angle A+\angle B=180^{\circ}\).
Let's use the transversal \(BD\). In \(\triangle ABD\) and \(\triangle CDB\), \(AB = CD\), \(AD = BC\) (Opposite sides of a parallelogram are equal) and \(BD=BD\) (Common side). So, \(\triangle ABD\cong\triangle CDB\) (SSS - Side - Side - Side congruence criterion).
\(\angle ADB=\angle CBD\) and \(\angle ABD=\angle CDB\) (Corresponding parts of congruent triangles are equal).
Since \(AD\parallel BC\) and \(BD\) is a transversal, \(\angle ADB\) and \(\angle CBD\) are alternate - interior angles. Similarly, for \(AB\parallel CD\) and \(BD\) as a transversal.
Now, \(\angle A+\angle B\):
We know that \(\angle A=\angle ADB+\angle ABD\) and \(\angle B=\angle CBD+\angle ABD\) (Angle - addition postulate). Since \(\angle ADB+\angle ABD+\angle CBD+\angle ABD = 180^{\circ}\) (Because \(AD\parallel BC\) and \(AB\) is a transversal, consecutive - interior angles \(\angle A\) and \(\angle B\) related to the transversal \(AB\) for parallel lines \(AD\) and \(BC\)).

Step3: Generalize

For any parallelogram \(ABCD\), if we consider two consecutive angles. Let the consecutive angles be \(\angle X\) and \(\angle Y\) where the sides forming the angles are such that one pair of opposite sides of the parallelogram is a transversal. Using the property of parallel lines (opposite sides of a parallelogram are parallel) and the consecutive - interior angles theorem (If \(l\parallel m\) and \(t\) is a transversal, then \(\angle1+\angle2 = 180^{\circ}\) where \(\angle1\) and \(\angle2\) are consecutive - interior angles), we can show that for consecutive angles \(\angle B\) and \(\angle A\) (with \(AD\parallel BC\) and \(AB\) as transversal) \(\angle B+\angle A=180^{\circ}\), and for \(\angle C\) and \(\angle B\) (with \(AB\parallel CD\) and \(BC\) as transversal) \(\angle C+\angle B = 180^{\circ}\)

Answer:

Statements	Reasons
2. Let \(BD\) be a transversal	Construction
3. \(\angle ABD=\angle CDB\), \(\angle ADB=\angle CBD\)	Alternate - interior angles (for \(AB\parallel CD\) and \(AD\parallel BC\) with transversal \(BD\))
4. \(\angle A=\angle ADB+\angle ABD\), \(\angle B=\angle CBD+\angle ABC\)	Angle - addition postulate
5. \(\angle A+\angle B=(\angle ADB+\angle ABD)+(\angle CBD+\angle ABC)\)	Substitution
6. \(\angle A+\angle B = 180^{\circ}\)	Since \(\angle ADB+\angle CBD\) and \(\angle ABD+\angle ABC\) form linear pairs (because of parallel lines \(AD\parallel BC\) and \(AB\) as transversal)
7. Similarly, \(\angle C+\angle B=180^{\circ}\)	Using \(AB\parallel CD\) and \(BC\) as transversal and angle - addition postulate