Perpendicular bisectors
Circumcenter of a Triangle Multiple proofs showing that a point is on a perpendicular bisector of a segment if and only if it is equidistant from the endpoints. Using this to establish the circumcenter, circumradius, and circumcircle for a triangle
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- Let's start off with segment AB, so that's point A,
- this is point B right over here,
- and lets set up a perpendicular bisector of this segment
- So it, it will be both perpendicular and it will split
- the segment in two, so thus we can call that line L,
- that's going to be a perpendicular, it's a perpendicular bisector,
- so it's gonna be, it'll intersect in a 90 degree angle
- and it bisects it
- This length and this length are equal,
- and we can even set, let's call this point right over here,
- let's call that M, maybe M for midpoint
- What I wanna prove first in this video,
- is that if we pick an arbitrary point on this line,
- that is a perpendicular bisector, off AB,
- 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 to the point,
- the same as that distance to the point,
- same as that distance from that point to B
- So let me pick an Arbitrary point from 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 since that 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
- Now, 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 the right angle, this is also right angled,
- This line is a perpendicular bisector of AB,
- and so we have two right triangles
- And if you 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, those are congruent
- So we can just use SAS, Side Angle Side congruency,
- Side Angle Side congruency
- So, we can write that triangle AMC, AMC is congruent,
- congruent to triangle BMC, to triangle BMC,
- by Side Angle Side congruency, Side Angle Side congruency,
- and so if both, if 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 wanna prove
- This arbitrary point C that sits on the perpendicular bisector of AB,
- is equidistant from both A and B,
- and I could've known that if I drew my C over here, or here,
- I would've 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, 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 like this, so this is my A,
- This is my B, and let's draw out some point, well call it C again,
- So let's say that's C right over here, and I'll, maybe I'll draw a C
- right down here
- So this is C and we're gonna start from the assumption
- that C is equidistant from A and B,
- so CA is going to be equal to CB,
- this is what we're gonna start of 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 AB
- So, we've drawn a triangle here, and we've done this before,
- And 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,
- if we draw it like this, so let's call,
- let's, let's just drop an altitude right over here,
- though we're really not dropping,
- 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 look like this,
- so that this was, so this was B, this is A and that C was up here,
- you could, you'd really be dropping this altitude,
- and so you can construct this line,
- so it, it has that, it is at a right angle with AB, and let me call
- this, the point in 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, this is the 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 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 means that AM,
- so that tells us that AM must be equal to BM,
- cause 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,
- it really is a 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 new, you know, just to review,
- we found, hey if any point sits on a perpendicular bisector
- of a segment, it's equidistant from the end points of a segment,
- And we went the other way, if any point is equidistant
- from the end points 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'm trying 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 point A, point B, and point C, we can 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,
- it will be perpen, actually, let me draw this a little different,
- coz 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
- Everytime I, okay, and then, let me draw it, let me,
- okay this one might be a little bit better,
- and we'll see which 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 perp,
- then draw the perpendicular bisector,
- so the perpendicular bisector might look
- something like that, might look something like that,
- and I don't want to make it necessarily intersecting C,
- coz 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 it's midpoint which,
- if I just roughly draw it, it looks
- like it's right over there, and then let me draw
- it's 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 now there's some interesting properties of point O,
- we know that since O sits on AB's perpendicular bisector,
- we know that the distant, the distance from O to B
- is going to be the distance from O to A
- That's what we proved in this first little proof over here,
- so we know, we know that OA, is going to be equal to OB,
- well that's kind of neat, but we also know that,
- coz 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 AC, that is 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,
- and 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, 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 BC's perpendicular,
- perpendicular bisector
- And what's neat about this simple little proof
- that we've set up on this video, is if we've shown,
- 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 O 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 this distance over here,
- will have a circle that goes through all of the vertices of B,
- oh so 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, circumradius,
- circumradius, and once again, we know we can construct it,
- Coz there is a point here, and it is centered at O, and this circle
- Because it goes through the vertices of our triangle,
- all of the vertices of our triangle, we say that it is circumscribe,
- circumspri, circum, I have trouble saying 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, circumscribed, about, about triangle ABC,
- which just means that, all three vertices lie on the circle,
- and that the circle has it, every point, is the circumradius away,
- from this circumcenter
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