The perpendicular bisector theorem

0:00- [Instructor] What I'd like to show in this video is that if I have a line segment... Let me just draw a line segment there. Let's say it's connecting the points A and B. Those are the end points of the line segment. And if I were to have a perpendicular bisector of this line segment, so that means a line that intersects it at a right angle, and where it intersects is the midpoint of AB.

0:24Let's just assume that that's the perpendicular bisector. So one way to think about it is this point over here, M, is the midpoint of the segment AB. So this distance is equal to this distance. AM is equal to MB. And I said it's perpendicular bisector. So this is a right angle, which of course means that this is a right angle as well.

0:46So what I'm going to prove first is if I have a perpendicular bisector like this, that any point on that perpendicular bisector is equidistant from the two endpoints of the segment. So if I were to pick some arbitrary point, let's call this point right over here, P, I wanna prove that P, and this will be true for any point because I picked P arbitrarily, that P is equidistant to A and to B.

1:17So let's do that. So first, let me construct two triangles. So I have triangle APM, and I have triangle BPM just like that. So I have these two triangles. Now, we know that AM is equivalent to MB. I already marked that there. And then of course we know that PM, which is in both of these triangles, is equal to itself.

1:46So we can now see that both of these triangles have a side angle and side in common. Side angle and side that are equal to each other. And so based on that, we can say... Let me write it over here. We can say that triangle APM, APM, is congruent to triangle BPM, BPM, by side, angle side.

2:22We see that, side, angle, side. Side, angle side. Now, why is that useful? Well, if we know these two triangles are congruent to each other, then the corresponding sides are congruent. So this length must be equivalent to this length. So we can say that AP is equal to BP.

2:41And so we've proven. We've proved it for an arbitrary point P, so it it's gonna be true for any point that I pick on this line, it's gonna be equidistant from the endpoint of that segment. Now I wanna prove it the other way around. Let's imagine a world where I have some point P that is equidistant from A and B.

3:06I want to prove that any point P like that is going to be on a perpendicular bisector. And so how do I do that? Well, let's once again draw some segment AB, A and B here, and we're gonna pick my arbitrary point P again. So let me pick my arbitrary point P right over there.

3:32And let's say that there is some midpoint M. Any segment's going to have some midpoint, so I can define that. So let's make that, M right over here. So we know some midpoint. So we know that AM is equal to MB, and we are assuming in this first, in the second proof that P is a point that is equidistant from A and B.

3:53So let me... Let me make it clear what we're proving. In the first proof, we wanted to prove that if P is on a perpendicular bisector of segment AB, then P is equidistant, equidistant from A and B. Well, now we're gonna prove the converse. We're gonna prove if P is equidistant, equidistant from A and B, then P is on perpendicular bisector of segment AB. So we're proving it the other way around.

4:48You're essentially swapping out the ifs and the thens here. So I've said that P is equidistant from A and B. So if I were to make segments between P and A, let me make one that's easier to see. So a segment that's between P and A and another segment that's between P and B, these are going to have the same lengths just like that.

5:14And so this is everything we now know. I've define... I define to some midpoint, any segment's going to have a midpoint. So AM is equal to MB. I'm picking some arbitrary point P that's equidistant from A and B, and that's why PA is equal to PB in length. And now let me draw the line that connects P and M.

5:34So if I... Well, that's fatter than I expected, but that's okay. All right. So I'll draw that line. And what I need to do is prove that that's the perpendicular bisector of segment AB. And to do that, I can once again go to triangle congruence because I have one side, set of sides that are congruent to each other.

5:55I have another set of side, and then this third side PM is common to both triangles. And so we know that triangle APM is congruent, is congruent to triangle BPM by side, side, side. And that means that all the corresponding angles are going to be equivalent. So that means that the...

6:24So we know from that that the measure of angle AMP, AMP is equal to the measure of angle BMP, BMP, because those are corresponding angles in two congruent triangles. And we also know that they're supplementary, and angle A, angle AMP and angle BMP are complimentary.

7:01Or another way of saying it, they both add up to 180 degrees. Well, if you have two things that are equal and they add up to 180 degrees, that means that either one of them, that the measure of angle AMP and the measure of angle A, BMP, BMP is equal to 90 degrees.

7:23So this has to be 90 degrees. These two angles are congruent to each other and they are supplementary. They add up to 180 degrees, so they each have to be 90. And just like that, we showed that PM, the line PM is indeed the perpendicular bisector. It contains M, so it's definitely bisecting at the midpoint and it's definitely perpendicular, which we just proved.