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

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