Question

given: abcd is a parallelogram. prove: ( mangle a + mangle b + mangle c + mangle d = 360^{circ} )

Explanation:

Step1: Analyze the first pair of same - side interior angles

In parallelogram \(ABCD\) with \(AD\parallel BC\) and \(AD\) as a transversal, \(\angle A\) and \(\angle D\) are same - side interior angles. By the property of same - side interior angles (when two parallel lines are cut by a transversal, same - side interior angles are supplementary), \(m\angle A + m\angle D=180^{\circ}\).

Step2: Analyze the second pair of same - side interior angles

Using \(BC\) as a transversal for \(AB\parallel DC\), \(\angle B\) and \(\angle C\) are same - side interior angles. So \(m\angle B + m\angle C = 180^{\circ}\).

Step3: Combine the two equations

We know that \(m\angle A+m\angle D = 180^{\circ}\) and \(m\angle B + m\angle C=180^{\circ}\). By the addition property of equality, if \(a = b\) and \(c = d\), then \(a + c=b + d\). Here \(a=m\angle A+m\angle D\), \(b = 180\), \(c=m\angle B+m\angle C\) and \(d = 180\). So \(m\angle A+m\angle D+m\angle B+m\angle C=180 + 180\).

Answer:

addition