+0

# https://web2.0calc.com/questions/not-sure-where-to-start-here

+1
444
3
+1438

Can anyone provide the answer for this?  TheXSquaredFactor was very helpful but I think it would make more sense if I could see the final product.

Thanks!

Jan 23, 2018

#1
+104807
0

What's the question  ???

Jan 23, 2018
#2
+1438
+1

A quadrilateral is called a "parallelogram" if both pairs of opposite sides are parallel. Show that if WXYZ is a parallelogram, then < W = < Y and < X = < Z.

(< = angle)

I put the link in the problem title.

AnonymousConfusedGuy  Jan 23, 2018
#3
+2344
+1

Go here https://web2.0calc.com/questions/not-sure-where-to-start-here to view my original response and the corresponding question.

I already explained in the previous post that opposite sides of a parallelogram are parallel by definition. Then, I encouraged you to try and figure out a relationship with two angles located on the same segment. In this case, $$\angle Z$$ and $$\angle Y$$ are same-side interior angles (sometimes referred to as consecutive interior angles). I know this because $$\overline{WZ}\parallel\overline{XY}$$, and $$\overline{ZY}$$ acts as the transversal.  Because there is proof of parallel segments, we also know that the same-side interior angles are supplementary. If the angles are known to be supplementary, then the sum of the angles equals 180°. By the definition of supplementary angles, then, $$m\angle Z+M\angle Y=180^{\circ}$$. Using the same logic as before, it is also possible to conclude that $$m\angle X+m\angle Y=180^{\circ}$$. Based on the two previous conclusions, both $$\angle Z$$ and $$\angle X$$ are supplementary to a common angle, $$\angle Y$$ in this case. When this occurs, the congruent supplements theorem states that $$\angle Z\cong\angle X$$. Utilize this logic again to prove that the other pair of opposite angles, $$\angle W\text{ and }\angle Y$$, are indeed congruent. The process is identical.

Jan 23, 2018