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## Perpendicular bisectors

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

# Circumcenter of a triangle

## 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.