Question

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

Explanation:

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

We know that \(m\angle A+m\angle D+m\angle B + m\angle C=(m\angle A + m\angle D)+(m\angle B + m\angle C)\). Since \(m\angle A + m\angle D = 180^{\circ}\) and \(m\angle B + m\angle C=180^{\circ}\), by the substitution property of equality \(m\angle A+m\angle D+m\angle B + m\angle C=180 + 180\).

Answer:

First blank: \(D\) (because when \(AD\) is a transversal for \(AB\parallel DC\), the angle that forms same - side interior angles with \(\angle A\) is \(\angle D\)). Second blank: supplementary (by the property of same - side interior angles for parallel lines). Third blank: \(DC\) (since \(AD\parallel BC\) and we are looking at \(\angle D\) and \(\angle C\), \(DC\) is the transversal). Fourth blank: substitution (we substitute \(m\angle A + m\angle D = 180\) and \(m\angle B + m\angle C = 180\) into \(m\angle A+m\angle D+m\angle B + m\angle C\)).