Proof: Trapezoid midsegment length

0:00- [Instructor] All right, what we wanna do in this video is prove that for a trapezoid, and remember, a trapezoid is just a quadrilateral that has a pair of parallel sides, and we call those the bases. And so, that would be BA and CD, that for a trapezoid like this, the midsegment that intersects the two legs of the trapezoid, those are the non-parallel sides in this example.

0:23It intersects the legs at their midpoint. And we could see that this length, CE, is equivalent to EB, and DF is equivalent to FA. We wanna prove that this midsegment's length is equal to the average of the lengths of the bases. Or another way to say it, we wanna say in this particular example, the length of EF, the length of EF is equal to 1/2 times the length of CD plus the length of BA.

0:56If you get inspired either at the beginning or at any time during this video, I encourage you to pause the video and see if you can complete the proof on your own. But now let's do it together. So, to do this proof I'm going to actually construct a bunch of triangles.

1:10So, I'm gonna create, let me put a new line here that goes from C through F and then just keeps going. And what essentially forms a new point of intersection if we were to extend BA further. So, let's extend BA further, and we call this new point right over here.

1:31Let's see, we already have A through F, so let's just call this one right over here, G. So, the first thing that I want to show is that this triangle up here and this triangle down here, that these two are congruent triangles. How can I do that? Well, we see that this angle is going to be equal to that angle because they are vertical angles.

1:53We also know that CD up here, if we, it is parallel to BA, or we could even say BG down here. And so, this angle and this angle would be alternate interior angles. And so, we already have two angles that are similar, or two angles that have the same measure. If two angles have the same measure, the third one is going to because they have to add up to 180.

2:20So, we already know that these are similar triangles. And then we also know that we have a side that is common between these two. This side FA is equal to DF, and so these are congruent triangles. Another way to think about it is, you know all the angles are the same and you have at least one corresponding side that is the same.

2:42That is enough to know that these are, these are congruent triangles. And so, we could say that triangle, triangle CDF, CDF, is congruent to triangle GAF, GAF. Now, why is this useful? Well, if they are congruent then we know that the lengths of corresponding sides are going to be the same.

3:11So, we know, for example, that the length of CD, let me put maybe three hash marks here, is going to be the same as this length of AG right over here. Or another way to think about it is we know that BG, this whole thing, is equal to CD because CD is the same thing as AG.

3:34So, it's equal, this right over here is the same thing as CD because it's congruent. BG is equal to CD plus BA, plus BA. This whole line length is just this length plus this length, and this length is CD. So, that's interesting. We already have something that looks a little bit like this.

3:57Now we just have to prove that EF is half of that. Well, to help us there, we'll continue to leverage the fact that these two triangles are congruent. So, we also know that this third corresponding side is also, also they have the same lengths. So, we could, let me, I guess I have to use four hashes now.

4:15So, that is equivalent to this one right over there. Now, why is this useful? Well, now I can prove the similarity, not the congruence, of two different triangles, but this is going to be very, very useful. Because I can show that this triangle right over here that I'm showing in red, the smaller one, triangle CEF, it is similar to triangle, maybe I'll do it in the slight blue color, CBA.

4:46How can I show that these are, actually CBG, I should say. How can I show that these are similar triangles? Well, what we need to show is that the ratio between at least two corresponding sides are the same. And maybe if we have an angle in between them, then that's, we have SAS similarity.

5:06And we can show that. The ratio between CE and CB is 1/2 because this is half of this whole thing because these two are the same. And we also know the ratio of CF to CG is 1/2. So, this side is to this whole side is the same as this side is to that whole side.

5:27And we know that the angle in between is the same for both triangles because that angle is shared by both triangles. And so, we could write by SAS, by Side Angle Side we know that triangle CEF is going to be similar to triangle CBG, CBG. Now, why is this useful?

5:57Well, as we said, you have the same, if we have two similar triangles the ratio between corresponding sides is going to be the same. And we see that the ratio between any side on CE and the corresponding side, or any side on CEF and the corresponding side on CBG, it's a ratio of 1/2.

6:16And so, one way to think about it is the ratio between EF, and I think you see how this is going. The ratio between EF to BG, which is this whole side right over here, which is the same thing as CD plus BA. So, the ratio of that to that is going to be the same as the ratio of CE to CB, or the ratio of CF to CG, which is equal to 1/2.

6:44Well, now I think you see how we get there. If we just multiply both sides by CD plus BA, and just as a reminder, that was the same thing as BG. If we multiply both sides times this, we get that original result that we wanted to prove.