If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:12:29

Video transcript

let's start off with segment a B 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 believe both perpendicular and it will split the segment in two so this we could call that line L that's going to be a perpendicular it's a perpendicular bisector so it's going to be it'll intersect at a 90 degree angle and it bisects it this length and this length are equal and we could even set let's call this point right over here let's call that M maybe M 4 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 a B then that arbitrary point will be an equal distant from a or the distance from that point to a will be the same as that distance to the point is the same as that this same is the distance from that point to B so let me pick an arbitrary point on this perpendicular bisector so let's call it 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 C a is equal to C B 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 a B and so we have two right triangles and if you don't even have to worry about that the right triangles if you look at triangle AMC you have this side is congruent to the corresponding side on triangle B MC then you have an angle in between that corresponds to this angle over here angle AMC corresponds to angle B MC and they're both 90 degrees so they are congruent and then you have the side MC that's on both triangles and those those are congruent so we can just use SAS side-angle-side congruence side-angle-side congruence so we can write that triangle a.m.c AMC is congruent congruent to triangle BMC to triangle BMC by side-angle-side congruence side-angle-side congruence and so if though they are congruent then all of the 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 a B 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 seed that sits on this line so that's fair enough so let me just write it so this means that AC is equal to is equal to BC now let's go the other way around let's say 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 draw out some point we'll call it C again so let's say that C right over here now maybe I'll draw see 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 C a is going to be equal to C B this is what we're going to start off with this is going to be our assumption and what we want to prove is is that C sits on the perpendicular bisector of the perpendicular bisector of a B 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 call let's let's just drop an altitude right over here I'll really not dropping it we're kind of lifting an altitude in this case but if you rotated this around so that the triangle look like this so that the triangle looked like this so that this was 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 a it is adding a right angle with a B 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 C M 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 a B 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 it's at a right angle and then we know that we know that cm is going to be equal to we know that cm is going to be equal to is going to be equal to itself and so we know by this is a right angle we have a leg and we have a hypotenuse we know by the rsh postulate rsh postulate rsh 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 a cm is congruent to triangle B cm by the RS h postulate well if they're congruent then the corresponding sides are going to be congruent so that means that am so that tells us that a M 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 a B so this line MC really is on the perpendicular bisector it really is part of 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 this was a neat you know just to review we found hey if any points it's 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's it on the perpendicular bisector of that segment so let's apply those ideas to a triangle down 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 it's a point a point B and Point C you could call this triangle ABC now let me just construct the perpendicular bisector of segment a B 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 it will be perpendicular this a little different because I the way I've drawn this triangle it's it's get 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 let me draw it a little bit every time line okay and then let me draw it let me okay this one might be a little bit better and we'll see what special case I was referring to so let's 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 a / to draw the perpendicular bisector so the perpendicular bisector might look something like that might look something like that and I don't want it to make it necessarily intersect and 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 take let me do the same thing for segment AC right over here let me take its midpoint which if I just roughly draw it looks like it's right over there and then let me draw its perpendicular bisector so it would look something like this 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 let's call that point just for fun let's call that point O and other some interesting properties of point O we know that since o sits on a B's perpendicular bisector we know that the distance from o to B is going to be the same as a from o to a that's what we proved in this first little proof over here so we know we know that o a is going to be equal to OB well that's kind of neat but we also know that because it's the intersection of this green perpendicular bisector and this yellow perpendicular bisector we also know because it sits on the perpendicular bisector of of of a a see that it's equidistant from a as it is to see so we know that Oh a is equal to OC now this is interesting Oh a is equal to OB Oh a 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 must be equal to OB OC must be equal to OB well if a point is equal sorry 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 it will look it will this definitely lies on DC's perpendicular perpendicular bisector and what's neat about the simple little proof that we've set up in this video is if 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 of 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 Oh a circumcenter circumcenter circum 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 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 B after 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 circumcircle and this distance right here we call it the circumradius circum radius and once again we know we can construct it because there is a point here and it is centered at O now this circle because it goes through the vertices of our triangle all of the vertices of our triangle we say that it is circumscribe circumcising it circumscribed about the triangle so we can say right over here that the circle o the circumcircle o so circle o circle o right over here is circumscribed circum described about about triangle ABC which just means that all three vertices lie on the circle and the circle as it's every point is the circum radius away from this circumcenter