### Video Transcript

Complete the following. If π΄π΅πΆπ· is a quadrilateral and the vector from π to π is equal to the vector from π to π and theyβre parallel, then the quadrilateral can always be classified as blank. Is it option (A) a trapezoid, option (B) a parallelogram, option (C) a kite, option (D) an isosceles trapezoid, or option (E) a rectangle?

In this question, weβre given a quadrilateral π΄π΅πΆπ·, and weβre told that the vector from π to π is equal to the vector from π to π. This means they have the same magnitude and direction, so we can conclude that theyβre parallel. We need to use this information to determine what type of quadrilateral π΄π΅πΆπ· is. And to determine this, we need to note weβre only really given one piece of information: the vector from π to π is exactly equal to the vector from π to π. And although this tells us they have the same magnitude and direction, thereβs a lot of different ways we could sketch these vectors.

For example, we could sketch one vector directly above the other, as shown. We could then note that this shape represents a square. However, it doesnβt necessarily need to be this way. For example, we couldβve drawn our vectors even further apart. Then we would have had a rectangle. But this is not the only change we couldβve made. We couldβve also drawn our vectors not one above the other. We couldβve drawn them separately in the plane. And sketching the remaining sides of quadrilateral π΄π΅πΆπ· gives us a shape which looks like a parallelogram.

However, we canβt answer this question just by looking at this graphically. We need to prove that this is a parallelogram. And to do this, we need to show that the opposite sides are parallel. We already know the side from π to π is parallel to the side from π to π. So we need to show this for the other two sides.

We can do this by using vectors. We can find an expression for the vector from π to π by following the vertices of our parallelogram. We get that the vector from π to π is equal to the vector from π to π added to the vector from π to π added to the vector from π to π. This is just an application of the triangle rule for the addition of vectors.

But now we can notice something interesting. We have ππ added to vector ππ, and weβre told in the question that vector ππ is equal to vector ππ. And the vector from π to π is exactly equal to the vector from π to π. However, we switch its direction, so the vector from π to π is negative the vector from π to π. So we can rewrite this as the vector from π to π added to the vector from π to π minus the vector from π to π.

But now we can use the fact that the vector from π to π is equal to the vector from π to π. We can replace this in our expression. And then we see the vector from π to π and subtracting the vector from π to π cancel to give the zero vector. So this simplifies to give us that the vector from π to π is equal to the vector from π to π. And if the two vectors are equal, this means they have the same magnitude and direction. And we canβt refine this any further. As weβve already shown, we couldβve had a rectangle, a square, or just a general parallelogram. Therefore, we were able to show if π΄π΅πΆπ· is a quadrilateral and the vector from π to π is equal to the vector from π to π, then quadrilateral π΄π΅πΆπ· is a parallelogram.