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Studying for a test? Prepare with these 8 lessons on Triangles.
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Video transcript
Let's start off with segment AB. So that's point A. This is point B right over here. And let's set up a perpendicular bisector of this segment. So it will be both perpendicular and it will split the segment in two. So thus we could call that line l. That's going to be a perpendicular bisector, so it's going to intersect at a 90-degree angle, and it bisects it. This length and this length are equal, and let's call this point right over here M, maybe M for midpoint. What I want to prove first in this video is that if we pick an arbitrary point on this line that is a perpendicular bisector of AB, then that arbitrary point will be an equal distant from A, or that distance from that point to A will be the same as that distance from that point to B. So let me pick an arbitrary point on this perpendicular bisector. So let's call that arbitrary point C. And so you can imagine we like to draw a triangle, so let's draw a triangle where we draw a line from C to A and then another one from C to B. And essentially, if we can prove that CA is equal to CB, then we've proven what we want to prove, that C is an equal distance from A as it is from B. Well, there's a couple of interesting things we see here. We know that AM is equal to MB, and we also know that CM is equal to itself. Obviously, any segment is going to be equal to itself. And we know if this is a right angle, this is also a right angle. This line is a perpendicular bisector of AB. And so we have two right triangles. And actually, we don't even have to worry about that they're right triangles. If you look at triangle AMC, you have this side is congruent to the corresponding side on triangle BMC. Then you have an angle in between that corresponds to this angle over here, angle AMC corresponds to angle BMC, and they're both 90 degrees, so they're congruent. And then you have the side MC that's on both triangles, and those are congruent. So we can just use SAS, side-angle-side congruency. So we can write that triangle AMC is congruent to triangle BMC by side-angle-side congruency. And so if they are congruent, then all of their corresponding sides are congruent and AC corresponds to BC. So these two things must be congruent. This length must be the same as this length right over there, and so we've proven what we want to prove. This arbitrary point C that sits on the perpendicular bisector of AB is equidistant from both A and B. And I could have known that if I drew my C over here or here, I would have made the exact same argument, so any C that sits on this line. So that's fair enough. So let me just write it. So this means that AC is equal to BC. Now, let's go the other way around. Let's say that we find some point that is equidistant from A and B. Let's prove that it has to sit on the perpendicular bisector. So let's do this again. So I'll draw it like this. So this is my A. This is my B, and let's throw out some point. We'll call it C again. So let's say that C right over here, and maybe I'll draw a C right down here. So this is C, and we're going to start with the assumption that C is equidistant from A and B. So CA is going to be equal to CB. This is what we're going to start off with. This is going to be our assumption, and what we want to prove is that C sits on the perpendicular bisector of AB. So we've drawn a triangle here, and we've done this before. We can always drop an altitude from this side of the triangle right over here. So we can set up a line right over here. Let me draw it like this. So let's just drop an altitude right over here. Although we're really not dropping it. We're kind of lifting an altitude in this case. But if you rotated this around so that the triangle looked like this, so this was B, this is A, and that C was up here, you would really be dropping this altitude. And so you can construct this line so it is at a right angle with AB, and let me call this the point at which it intersects M. So to prove that C lies on the perpendicular bisector, we really have to show that CM is a segment on the perpendicular bisector, and the way we've constructed it, it is already perpendicular. We really just have to show that it bisects AB. So what we have right over here, we have two right angles. If this is a right angle here, this one clearly has to be the way we constructed it. It's at a right angle. And then we know that the CM is going to be equal to itself. And so this is a right angle. We have a leg, and we have a hypotenuse. We know by the RSH postulate, we have a right angle. We have one corresponding leg that's congruent to the other corresponding leg on the other triangle. We have a hypotenuse that's congruent to the other hypotenuse, so that means that our two triangles are congruent. So triangle ACM is congruent to triangle BCM by the RSH postulate. Well, if they're congruent, then their corresponding sides are going to be congruent. So that tells us that AM must be equal to BM because they're their corresponding sides. So this side right over here is going to be congruent to that side. So this really is bisecting AB. So this line MC really is on the perpendicular bisector. And the whole reason why we're doing this is now we can do some interesting things with perpendicular bisectors and points that are equidistant from points and do them with triangles. So just to review, we found, hey if any point sits on a perpendicular bisector of a segment, it's equidistant from the endpoints of a segment, and we went the other way. If any point is equidistant from the endpoints of a segment, it sits on the perpendicular bisector of that segment. So let's apply those ideas to a triangle now. So let me draw myself an arbitrary triangle. I'll try to draw it fairly large. So let's say that's a triangle of some kind. Let me give ourselves some labels to this triangle. That's point A, point B, and point C. You could call this triangle ABC. Now, let me just construct the perpendicular bisector of segment AB. So it's going to bisect it. So this distance is going to be equal to this distance, and it's going to be perpendicular. So it looks something like that. And it will be perpendicular. Actually, let me draw this a little different because of the way I've drawn this triangle, it's making us get close to a special case, which we will actually talk about in the next video. Let me draw this triangle a little bit differently. OK. This one might be a little bit better. And we'll see what special case I was referring to. So this is going to be A. This is going to be B. This is going to be C. Now, let me take this point right over here, which is the midpoint of A and B and draw the perpendicular bisector. So the perpendicular bisector might look something like that. And I don't want it to make it necessarily intersect in C because that's not necessarily going to be the case. But this is going to be a 90-degree angle, and this length is equal to that length. And let me do the same thing for segment AC right over here. Let me take its midpoint, which if I just roughly draw it, it looks like it's right over there. And then let me draw its perpendicular bisector, so it would look something like this. So this length right over here is equal to that length, and we see that they intersect at some point. Just for fun, let's call that point O. And now there's some interesting properties of point O. We know that since O sits on AB's perpendicular bisector, we know that the distance from O to B is going to be the same as the distance from O to A. That's what we proved in this first little proof over here. So we know that OA is going to be equal to OB. Well, that's kind of neat. But we also know that because of the intersection of this green perpendicular bisector and this yellow perpendicular bisector, we also know because it sits on the perpendicular bisector of AC that it's equidistant from A as it is to C. So we know that OA is equal to OC. Now, this is interesting. OA is equal to OB. OA is also equal to OC, so OC and OB have to be the same thing as well. So we also know that OC must be equal to OB. OC must be equal to OB. Well, if a point is equidistant from two other points that sit on either end of a segment, then that point must sit on the perpendicular bisector of that segment. That's that second proof that we did right over here. So it must sit on the perpendicular bisector of BC. So if I draw the perpendicular bisector right over there, then this definitely lies on BC's perpendicular bisector. And what's neat about this simple little proof that we've set up in this video is we've shown that there's a unique point in this triangle that is equidistant from all of the vertices of the triangle and it sits on the perpendicular bisectors of the three sides. Or another way to think of it, we've shown that the perpendicular bisectors, or the three sides, intersect at a unique point that is equidistant from the vertices. And this unique point on a triangle has a special name. We call O a circumcenter. And because O is equidistant to the vertices, so this distance-- let me do this in a color I haven't used before. This distance right over here is equal to that distance right over there is equal to that distance over there. If we construct a circle that has a center at O and whose radius is this orange distance, whose radius is any of these distances over here, we'll have a circle that goes through all of the vertices of our triangle centered at O. So our circle would look something like this, my best attempt to draw it. And so what we've constructed right here is one, we've shown that we can construct something like this, but we call this thing a circumcircle, and this distance right here, we call it the circumradius. And once again, we know we can construct it because there's a point here, and it is centered at O. Now this circle, because it goes through all of the vertices of our triangle, we say that it is circumscribed about the triangle. So we can say right over here that the circumcircle O, so circle O right over here is circumscribed about triangle ABC, which just means that all three vertices lie on this circle and that every point is the circumradius away from this circumcenter.