What I want to do in this video is prove
that the circumcenter of a right triangle,
is actually the midpoint of the hypotenuse, and to do that,
I'm gonna take, first take a look at the perpendicular bisector
of one of the legs, of this, of this right triangle
So, let me construct the perpendicular bisector
of leg BC right over here, so it's going to look something like this,
it's going to look something like this, this,
it intersects at a right angle, its perpendicular, and it bisects it
So B, this is from B to this point
which we'll call M, maybe M for midpoint,
is the same, as it is from M to C,
so those two distances are going to be equal,
and let's call the point where this perpendicular bisector
intersects the hypotenuse,
let's call this O,and we're gonna prove that O
is the circumcenter of this right triangle
Now, the first thing that you might realize,
and this is what we've seen in many problems,
the triangle OBM, looks similar to triangle ABC,
and its actually not too hard to prove,
they both already have a 90 degree angle,
so if we show that they, they both have another angle
Another set of corresponding angles
that are congruent to each other, then we know that they're similar
By AA similarity, and they both clearly share this angle,
right over here, OBC is part of the smaller triangle,
and ABC which is really the same angle,
is part of the larger triangle, and so, and they also obviously share
a 90 degree angle, so by AA triangle similarity,
we have triangle OBM, OBM is similar, is similar to triangle ABC,
is similar to triangle ABC, and what's useful about this?
Is we know similar triangles are ratios between corresponding sides
are constant, so for example, we know that the ratio between side BM,
which is on the smaller triangle, we know that the ratio between BM
Let me do this in a different color, just to, just for the sake of it,
we know that the ratio between BM
and BC, BM and BC, the ratio of this side on the smaller
triangle to the corresponding side on the larger triangle
Is going to be the same as the ratio of the hypotenuse
on the smaller triangle, BO to the hypotenuse of the larger triangle,
because they are similar, well we know what the ratio of BM to BC is,
BM is half of BC, so this ratio over here
is going to be equal to one half, this is M is the bi,
of the midpoint of these things,
so this is exactly the same distance as this,
so this is one half of the entire BC,
so if one half is equal to BM, over BC is equal to BO over BA
We then know, if we just kind of ignore this middle part,
right over here, that one half is equal to BO over BA,
over BA, if you cross multiply it,
if you cross multiply, you see that, well there's multiple ways
to think about, but you could just cross multiply, and you say BA,
is equal to 2BO, or if you divide both sides by two,
and their really equivalent statements one half BA is equal to BO,
so BO is one half of BA, so this is one half BA,
and so this other length, AO right over here
This is going to be B, this is going to be, this going to be BA,
minus one half BA, so this is also going to be, one half BA,
and so, this segment right over here,
AO, AO is going to be congruent to OB
So what we just shown, first of all,
is that this perpendicular bisector, right over here,
The perpendicular bisector of segment BC,
it intersects the hypotenuse of our right triangle at the midpoint,
So we've already established,
so we, one thing that we've already established,
is O, is the midpoint, is the midpoint,
of the hypotenuse, of the hypotenuse, of the hypotenuse AB,
well, that by itself is interesting,
but, we also know that if a point sits on a perpendicular bisector
of a segment, is equidistant, it's equal distant
from the end point of the segment, we'd show that in a previous video
So we also know that O, OB that's equidistant to the end points
of the segment, right over here, that OB is equal to OC, but we know,
from this first statement right over here,
that OB is alsoequal to OA, OB is also equal to OA,
its of OB is equal to OC
OB is equal to OA, that means OC must be equal to OA,
OC must be equal to OA
Or another way to think about it, is at this point O,
Is equal distant from all of the points on our tri,
all of the vertices, I should say, this point O is equidistant,
from all of the vertices of our triangle, of our triangle,
So this distance, this distance, which is really going to become
our circumradius, is the same as this distance right over here,
which is the same as this distance right over there
So that we know that O is equidistant,
equidistant to all, all vertices, which is another way of saying
that O is the circumcenter, O is the circumcenter, so we've just
proven that if you have the circumcenter of a right triangle,
it is the midpoint of the hypotenuse of the right triangle,
or the other way around, that the hypotenuse of the right triangle
is the circumcenter, because you only have one circumcenter of any,
of any triangle