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Sal proves that a point is the midpoint of a segment using triangle congruence. Created by Sal Khan.
Video transcript
So we have these two parallel lines, line segment AB and line segment CD. I should say they are parallel line segments. And then we have these transversals that go across them. So you have this transversal BC right over here. And you have this transversal AD. And what this diagram tells us is that the distance between A and E-- this little hash mark-- says that this line segment is the same distance as the distance between E and D. Or another way to think about it is that point E is at the midpoint, or is the midpoint, of line segment AD. And what I want to think about in this video is, is point E also the midpoint of line segment BC? So this is the question right over here. So is E the midpoint of line segment BC? And you could imagine, based on a lot of the videos we've been seeing lately, maybe it has something to do with congruent triangles. So let's see if we can set up some congruency relationship between the two obvious triangles in this diagram. We have this triangle up here on the left. And then we have this triangle down here. This one kind of looks like it's pointing up. This one looks like it's pointing down. So there's a bunch of things we know about vertical angles and angles of transversals. The most obvious one is that we have this vertical. We know that angle AEB is going to be congruent. Or its measure is going to be equal to the measure of angle CED. So we know that angle AEB is going to be congruent to angle DEC, which really just means they have the exact same measure. And we know that because they are vertical angles. Now, we also know that AB and CD are parallel. So this line right over here, this is a transversal. And there's actually several ways that we can do this problem. But we know that this is a transversal. And there's a couple of ways to think about it right over here. So let me just continue the transversal, so we get to see all of the different angles. You could say that this angle right here, angle ABE-- so this is its measure right over here-- you could say that it is the alternate interior angle to angle ECD, to this angle right over there. And if that didn't jump out at you, you would say that the corresponding angle to this one right over here is this angle right up here. If you were to continue this line off a little bit, these are the corresponding angles. And then this one is vertical. But either way, angle ABE-- let me be careful. Angle ABE is going to be congruent to angle DCE. And we could say because it's alternate interior angles. I'll just write a little code here. So Alt interior angles. And then we have an interesting relationship. We have an angle congruent to an angle, another angle congruent to an angle. And then the next side is congruent to the next side over here. So pink, green, side. Pink, green, side. So we can employ AAS, angle-angle-side. And it's in the right order. So now, we know that triangle-- we have to make sure that we get the letters right here, that we have the right corresponding vertices. We can say that triangle AEB-- actually, let me start with the angle just to make it interesting. Angle BEA, so we're starting with the magenta angle, going to the green angle, and then going to the one that we haven't labeled. So angle BEA, we can say, is congruent to angle-- we start with the magenta vertices-- C, go to the center, E, and then go the unlabeled one, D. And we know this because of angle-angle-side. And they correspond to each other-- magenta-green-side, magenta-green-side. They're all congruent. So this is from AAS. And then, if we know that they are congruent, then that means corresponding sides are congruent. So then we know these two triangles are congruent. So that means that their corresponding sides are congruent. So then we know that length of BE is going to be equal-- and that's the segment that's between the magenta and the green angles. The corresponding side is side CE between the magenta and the green angles-- is equal to CE. And this just comes out of the previous statement. If we number them, that's 1, that's 2, and that's 3. And so that comes out of statement 3. And so we have proven this. E is the midpoint of BC. It comes straight out of the fact that BE is equal to CE. So I can mark this off with hash. This line segment right over here is congruent to this line segment right over here, because we know that those two triangles are congruent. And I've inadvertently, right here, done a little two-column proof. This over here on the left-hand side is my statement. And then on the right-hand side, I gave my reason. And we're done.